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levou 18 s
Como pensou
**Parsing trigonometric equations**
I need to parse the equation which involves some trigonometric identities. Starting with a base of 0.5 for the cosine and sine functions, I realize the domain should be between 0 and 1 because cosine’s maximum is 1. The possible solutions for theta are pi/2 + 2kpi or pi + 2kpi. Calculating the angles leads to candidates like pi/8, pi/4, and 5pi/8, but I need to see which ones respect the constraints of the equation. Overall, I want to clarify the reasoning around the log domain and squaring possibilities.
Para que o logaritmo exista, sua base deve ser positiva e diferente de 1, e seu argumento deve ser positivo.
Como \log_b a=1 implica a=b, temos
0{,}5\sen(4x)=0{,}5\cos(4x)+0{,}5,
ou seja,
\sen(4x)-\cos(4x)=1.
Usando \sen t-\cos t=\sqrt2\,\sen(t-\pi/4), obtemos
\sen\left(4x-\frac{\pi}{4}\right)=\frac1{\sqrt2}.
Logo,
4x=\frac{\pi}{2}+2k\pi
\quad\text{ou}\quad
4x=\pi+2k\pi,\qquad k\in\mathbb Z.
No intervalo x\in[0,\pi], os candidatos são
x=\frac{\pi}{8},\ \frac{5\pi}{8},\ \frac{\pi}{4},\ \frac{3\pi}{4}.
- Para x=\frac{\pi}{8} ou \frac{5\pi}{8}, a base e o argumento são ambos \frac12. Portanto, o logaritmo vale 1.
- Para x=\frac{\pi}{4} ou \frac{3\pi}{4}, a base e o argumento são ambos 0, e o logaritmo não está definido.
Assim, há 2 valores de x que satisfazem a equação.
Resposta: B
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levou 14 s
Como pensou
Para que o logaritmo exista, sua base deve ser positiva e diferente de 1, e seu argumento deve ser positivo.
Como \log_b a=1 implica a=b, temos
0{,}5\sen(4x)=0{,}5\cos(4x)+0{,}5
\quad\Longrightarrow\quad
\sen(4x)-\cos(4x)=1.
Usando a identidade
\sen(4x)-\cos(4x)=\sqrt2\,\sen\left(4x-\frac{\pi}{4}\right),
obtemos
4x=\frac{\pi}{2}+2k\pi
\quad\text{ou}\quad
4x=\pi+2k\pi,\qquad k\in\mathbb Z.
Como x\in[0,\pi], temos 4x\in[0,4\pi]. Portanto, os candidatos são
x=\frac{\pi}{8},\quad \frac{\pi}{4},\quad
\frac{5\pi}{8},\quad \frac{3\pi}{4}.
- Para x=\frac{\pi}{4} ou \frac{3\pi}{4}, temos \cos(4x)=-1, de modo que a base é 0. Esses valores são inválidos.
- Para x=\frac{\pi}{8} ou \frac{5\pi}{8}, a base e o argumento são ambos \frac12, e \log_{1/2}(1/2)=1.
Logo, existem 2 valores de x.
Resposta: B
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levou 9 s
Como pensou
**Solving the log equation**
I need to parse the equation: log base (0.5 cos(4x)+0.5) of (0.5 sin(4x)) = 1. To solve for x in [0, pi], I know that if log_b(argument) = 1, then argument must equal b. I simplify this to 0.5 sin(4x) = 0.5 cos(4x) + 0.5 and find that t = 4x must meet the conditions for valid bases and arguments. I’ll solve for possible values of t and determine the corresponding x values. Two solutions emerge: x = pi/8 and 5pi/8.**Explaining the log equivalence**
I need to explain the equivalence of the logarithm equals 1 carefully, possibly in Portuguese. It’s important to note that t = pi + 2kpi is invalid in this context. I’ll count and clarify any equation notation, particularly focusing on the log base subscript. I want to ensure that any solutions for t are correctly derived using sine and cosine. Finally, I’ll wrap up with a conclusion that summarizes my findings and ensures the explanation is clear.
Defina t=4x. Como \log_b(a)=1 exige a=b, a equação implica
\frac12\sen t=\frac12\cos t+\frac12,
ou seja,
\sen t-\cos t=1.
Usando \sen t-\cos t=\sqrt2\sen(t-\pi/4), obtemos, em 0\le t\le4\pi,
t=\frac{\pi}{2},\quad \pi,\quad \frac{5\pi}{2},\quad 3\pi.
Os valores t=\pi e t=3\pi não servem: o argumento do logaritmo é zero (e a base também é zero). Restam t=\pi/2 e t=5\pi/2, que correspondem a x=\pi/8 e x=5\pi/8.
Há, portanto, 2 valores de x.
Resposta: B
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levou 13 s
Como pensou
I need to solve log base (0.5cos4x + 0.5) of (0.5 sin4x) = 1, requiring the base to be positive and not 1, and the argument positive. Setting the argument equal to the base gives sin4x − cos4x = 1, which simplifies to √2 sin(4x − π/4) = 1, so sin(4x − π/4) = √2/2.
This gives 4x = π/2 + 2kπ or 4x = π + 2kπ. The second case makes sin4x = 0, invalidating the argument, but the first case gives base 0.5 and argument 0.5, both valid. Solving x = π/8 + kπ/2 within [0, π] yields two solutions: π/8 and 5π/8, so the answer is B.
Condições de existência: base 0{,}5\cos(4x)+0{,}5>0 e \neq 1; argumento 0{,}5\,\text{sen}(4x)>0.
Resolvendo a equação:
0{,}5\,\text{sen}(4x)=0{,}5\cos(4x)+0{,}5 \;\Rightarrow\; \text{sen}(4x)-\cos(4x)=1
\sqrt{2}\,\text{sen}\left(4x-\tfrac{\pi}{4}\right)=1 \;\Rightarrow\; \text{sen}\left(4x-\tfrac{\pi}{4}\right)=\tfrac{\sqrt2}{2}
Logo 4x-\frac{\pi}{4}=\frac{\pi}{4}+2k\pi ou 4x-\frac{\pi}{4}=\frac{3\pi}{4}+2k\pi, isto é:
- 4x=\frac{\pi}{2}+2k\pi: aqui \text{sen}(4x)=1, \cos(4x)=0, base =0{,}5 (válida, \neq1) e argumento =0{,}5>0. Válido.
- 4x=\pi+2k\pi: aqui \text{sen}(4x)=0, argumento nulo. Inválido.
Soluções em [0,\pi]: x=\frac{\pi}{8}+\frac{k\pi}{2} → x=\frac{\pi}{8} e x=\frac{5\pi}{8} (o próximo, \frac{9\pi}{8}, excede \pi).
Portanto, há 2 valores de x.
Resposta: B
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levou 10 s
Como pensou
Setting the base as 0.5cos4x+0.5 and argument as 0.5sin4x, with constraints base>0, base≠1, argument>0, I simplify the equality to sin4x−cos4x=1, which becomes √2 sin(4x−π/4)=1, giving sin(4x−π/4)=√2/2 and solving 4x−π/4=π/4+2kπ.
Checking both resulting cases, only 4x=π/2+2kπ satisfies the validity constraints (sin4x=1, cos4x=0), while the other case fails since sin4x=0. Solving x=π/8+kπ/2 within [0,π] gives two solutions: π/8 and 5π/8.
Resposta: (B) 2
Condição de existência. A base é b = 0{,}5\cos(4x) + 0{,}5 e o logaritmando é a = 0{,}5\,\text{sen}(4x). Precisamos de a > 0, b > 0 e b \neq 1.
Resolução. Como \log_b a = 1, temos a = b:
0{,}5\,\text{sen}(4x) = 0{,}5\cos(4x) + 0{,}5 \;\Rightarrow\; \text{sen}(4x) - \cos(4x) = 1.
Escrevendo o lado esquerdo como um único seno:
\sqrt{2}\,\text{sen}\!\left(4x - \tfrac{\pi}{4}\right) = 1 \;\Rightarrow\; \text{sen}\!\left(4x - \tfrac{\pi}{4}\right) = \tfrac{\sqrt{2}}{2}.
Isso dá duas famílias de soluções:
-
4x - \tfrac{\pi}{4} = \tfrac{\pi}{4} + 2k\pi, ou seja, 4x = \tfrac{\pi}{2} + 2k\pi.
Então \text{sen}(4x) = 1 e \cos(4x) = 0.
A base vale 0{,}5 e o logaritmando vale 0{,}5. As condições são satisfeitas, e a família é válida.
-
4x - \tfrac{\pi}{4} = \tfrac{3\pi}{4} + 2k\pi, ou seja, 4x = \pi + 2k\pi.
Então \text{sen}(4x) = 0, o logaritmando é nulo e a família é inválida.
Contagem em [0, \pi]. Da família válida, x = \tfrac{\pi}{8} + \tfrac{k\pi}{2}. No intervalo, isso dá x = \tfrac{\pi}{8} e x = \tfrac{5\pi}{8}, ou seja, 2 soluções.
Resposta: B
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levou 7 s
Como pensou
I need log base b of a equal to 1, meaning a = b with b positive and not 1. Setting 0.5 sin4x equal to 0.5 cos4x + 0.5 gives sin4x − cos4x = 1, which I rewrite as √2 sin(4x − π/4) = 1, so sin(4x − π/4) = √2/2, leading to two families of solutions for 4x − π/4.
Solving these gives x = π/8 + kπ/2 or x = π/4 + kπ/2. Restricting to [0, π], the candidates are π/8, 5π/8, π/4, 3π/4. Checking validity, π/4 and 3π/4 make sin4x = 0 so a = 0, which is invalid for a logarithm base, leaving only π/8 and 5π/8 as valid solutions — so there are 2 solutions.
Condições de existência. Para que log₍b₎(a) = 1 faça sentido, é preciso ter b > 0, b ≠ 1 e a > 0, em que
- b = 0,5·cos(4x) + 0,5,
- a = 0,5·sen(4x).
Resolução. log₍b₎(a) = 1 equivale a a = b:
0,5·sen(4x) = 0,5·cos(4x) + 0,5, ou seja, sen(4x) − cos(4x) = 1.
Escrevendo sen(4x) − cos(4x) = √2·sen(4x − π/4), obtemos
sen(4x − π/4) = √2/2.
Isso dá duas famílias de soluções:
- 4x − π/4 = π/4 + 2kπ, logo 4x = π/2 + 2kπ e x = π/8 + kπ/2;
- 4x − π/4 = 3π/4 + 2kπ, logo 4x = π + 2kπ e x = π/4 + kπ/2.
Soluções em [0, π].
- Primeira família: x = π/8 e x = 5π/8.
- Segunda família: x = π/4 e x = 3π/4.
Verificação das condições.
- x = π/8 e x = 5π/8: 4x = π/2 ou 5π/2, então sen(4x) = 1 e cos(4x) = 0. Assim a = b = 0,5, que é positivo e diferente de 1. Servem.
- x = π/4 e x = 3π/4: 4x = π ou 3π, então sen(4x) = 0 e a = 0. Isso viola a > 0. Não servem.
Portanto, restam 2 valores de x.
Resposta: B
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levou 2 min 35 s
Como pensou
The problem is in Portuguese and concerns an equation.
The equation simplifies to 0.5 sin(4x) = 0.5 cos(4x) + 0.5. I am substituting θ = 4x and checking the domain conditions for the logarithm.
The equation sin θ − cos θ = 1 is rewritten as √2 sin(θ − π/4) = 1. The general solutions are θ = π/2 + 2kπ and θ = π + 2kπ.
The solutions are verified. θ = π/2 + 2kπ gives sin θ = 1 and cos θ = 0, satisfying the equation. θ = π + 2kπ gives sin θ = 0 and cos θ = −1, also satisfying it. Domain checks for the logarithm argument and base are now being applied.
The case θ = π + 2kπ is invalid because the logarithm argument becomes zero. Only θ = π/2 + 2kπ remains valid. The base 0.5 satisfies the domain conditions.
The interval [0, 4π] contains only two valid values of θ: π/2 and 5π/2. The third candidate 9π/2 exceeds the upper bound. I am now verifying that both values satisfy all domain conditions.
The two valid values of θ map to x = π/8 and x = 5π/8. Both lie in the interval [0, π]. No additional solutions exist.
I verified that b = 1 is impossible because a would also need to equal 1, which is impossible since sin θ cannot exceed 1. I also solved the equation by squaring both sides and found only sin θ = 0 or sin θ = 1. This confirms the earlier result of exactly two valid solutions.
The base and argument simplify to cos²(2x) and sin(2x)cos(2x). The equation becomes log_{cos²(2x)} (sin(2x)cos(2x)) = 1. This implies sin(2x)cos(2x) = cos²(2x).
The equation sin(2x)cos(2x) = cos²(2x) is factored into cos(2x)[sin(2x) − cos(2x)] = 0. Since a zero base is invalid, only tan(2x) = 1 remains. This yields x = π/8 + kπ/2, which maps to θ = π/2 + 2kπ. The result matches the earlier derivation.
The case where both the argument and base equal zero is being reconsidered. When cos(2x) = 0, the logarithm base becomes zero and the argument also becomes zero. This raises a question about whether such points should be counted as valid solutions.
The base is never equal to 1 because sin²(2x) + cos²(2x) = 1 forbids both values from being ±1 simultaneously. The argument is strictly positive since it equals sin²(2x). These checks confirm that only the two valid solutions remain.
The interval is confirmed as [0, π]. The answer is 2, which corresponds to option B.
A condição \log_b a = 1 equivale a a = b, desde que o logaritmo esteja definido, isto é, a > 0, b > 0 e b \neq 1.
Substituindo as expressões dadas, obtém-se
\operatorname{sen}(4x) = \cos(4x) + 1.
As soluções dessa equação trigonométrica são
4x = \frac{\pi}{2} + 2k\pi \quad\text{ou}\quad 4x = \pi + 2k\pi, \quad k \in \mathbb{Z}.
No intervalo x \in [0,\pi] (equivalentemente 4x \in [0,4\pi]) os candidatos são
x = \frac{\pi}{8},\ \frac{5\pi}{8},\ \frac{\pi}{4},\ \frac{3\pi}{4}.
Nos dois últimos valores tem-se argumento nulo e base nula, o que torna o logaritmo indefinido. Nos dois primeiros valores a base vale 1/2 e o argumento vale 1/2, de modo que todas as condições de existência são satisfeitas e a igualdade original se verifica.
(O mesmo conjunto de soluções válidas é recuperado reescrevendo a base como \cos^2(2x) e o argumento como \operatorname{sen}(2x)\cos(2x) e impondo as restrições de domínio.)
Portanto existem exatamente dois valores de x.
Resposta: B
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levou 20 s
Como pensou
We need answer in Portuguese. Need solve. Let's restate equation:
log_{(0,5·cos(4x)+0,5)}(0,5·sen(4x)) = 1.
Need number of x in [0, π] satisfying.
We need solve. Let base = 0.5 cos(4x)+0.5 = (cos(4x)+1)/2 = cos^2(2x)? Since cos4x = 2cos^2 2x -1, +1 = 2cos^2 2x, times 0.5 = cos^2(2x). Yes base = cos^2(2x). Argument = 0.5 sin(4x) = sin(4x)/2 = sin(2x) cos(2x)? Since sin4x = 2 sin2x cos2x, half = sin2x cos2x. So argument = sin(2x) cos(2x).
Log equation: log_{b}(a)=1 => a = b, with base b >0, b≠1, a>0. Here base b = cos^2(2x). For log defined: b>0, b≠1, a>0. b = cos^2(2x) ≥0. b>0 means cos(2x) ≠0. b≠1 means cos^2(2x) ≠1 => cos(2x) ≠ ±1 => sin(2x) ≠0? Actually cos^2=1 => cos(2x)=±1 => 2x = kπ => x = kπ/2. But base = 1 invalid. Also a>0: sin(2x) cos(2x) >0 => (1/2) sin(4x)>0 => sin(4x)>0.
Equation a=b: sin(2x) cos(2x) = cos^2(2x). Let c = cos(2x), s = sin(2x). c ≠0, c^2 ≠1, s c >0. Equation s c = c^2 => c(s - c)=0. Since c≠0, s = c. Also s c >0? If s=c, then c^2>0 true for c≠0. Base validity c^2≠1 => c ≠ ±1. So need sin(2x)=cos(2x), and cos(2x)≠0, ±1. sin(2x)=cos(2x) => tan(2x)=1 if cos≠0. So 2x = π/4 + kπ? Wait tan=1 => 2x = π/4 + kπ, but also sin=cos positive? Let's solve: sinθ = cosθ => θ = π/4 + kπ? Actually sinθ=cosθ => tanθ=1 => θ = π/4 + kπ. At these θ, sin and cos both positive or both negative? For θ=π/4 + kπ, sin=cos = ±√2/2. Product positive? If k even, positive; if k odd, negative. Argument positive condition sinθ cosθ >0 means product positive => k even. Also θ=2x in [0,2π] since x∈[0,π] => 2x∈[0,2π]. Need θ = π/4 + kπ, k integer, within [0,2π]. Possible θ: π/4 (k=0), 5π/4 (k=1). But for k=1, sin=cos=-√2/2, product positive? (-)(-) positive yes. Wait sin and cos both negative? At 5π/4, sin=-√2/2, cos=-√2/2, product positive. So argument positive yes. But check base: cos^2(5π/4)=1/2 valid. Equation log_{1/2}(0.5 sin(5π/2?) Wait sin(4x)=sin(2θ)=sin(5π/2)=sin(π/2)=1? Let's compute argument = 0.5 sin(4x) = 0.5 sin(2θ). For θ=5π/4, 2θ=5π/2, sin(5π/2)=1, argument=0.5. Base = cos^2(2x)=cos^2(5π/4)=1/2. log_{1/2}(0.5)=1 yes. So k=1 valid. Also k=0: θ=π/4, 2θ=π/2, sin=1, argument=0.5, base=cos^2(π/4)=1/2, log=1. So two solutions? But wait maybe other cases due to base? Let's carefully analyze original equation.
Equation: log_{cos^2(2x)}(sin(2x)cos(2x)) = 1.
Let t = 2x. x∈[0,π] => t∈[0,2π]. Equation log_{cos^2 t}(sin t cos t) = 1.
Domain: base = cos^2 t >0 and ≠1. So cos t ≠0, ±1. Argument = sin t cos t >0.
Equation => sin t cos t = cos^2 t. Since cos^2 t >0, divide by cos^2 t: tan t = 1? Wait sin t cos t / cos^2 t = tan t = 1, provided cos t ≠0. So tan t = 1. t = π/4 + kπ. Domain: cos t ≠0, ±1. For t = π/4 + kπ, cos t = ±√2/2, not 0 or ±1. So okay. Argument positive: sin t cos t = (1/2) sin(2t) >0. For t = π/4 + kπ, 2t = π/2 + 2kπ. sin(2t)=sin(π/2 + 2kπ)=1 >0 always. Wait earlier I computed sin(2θ)? Let's recalc: Argument = sin t cos t = 0.5 sin(2t). For t=5π/4, 2t=5π/2 = π/2 + 2π? Actually 5π/2 = 2π + π/2, sin=1. So positive. So both k=0,1 in [0,2π]? t=π/4, 5π/4. Also maybe t=9π/4? But [0,2π] includes only π/4 and 5π/4. So 2 solutions. But options include B)2. Is it that simple? Wait check base definition: log_{(0,5·cos(4x)+0,5)}(0,5·sen(4x)) = 1. Base = 0.5 cos(4x)+0.5. Let's express in t=2x: cos(4x)=cos(2t)=2cos^2 t -1. Base = 0.5(2cos^2 t -1)+0.5 = cos^2 t. Yes.
Argument = 0.5 sin(4x)=0.5 sin(2t)= sin t cos t. Yes.
Equation log_{cos^2 t}(sin t cos t)=1 => sin t cos t = cos^2 t. Domain: cos^2 t >0, ≠1, argument >0. Equation equivalent to sin t cos t = cos^2 t. Divide by cos t? Wait if cos t ≠0, divide by cos^2 t: tan t =1. But what if cos^2 t? yes. However, also consider possibility cos t =0? Domain excludes base=0. cos^2 t =0 => base=0 invalid. So no.
But wait, log equation: log_b(a)=1 means a=b^1=b. Yes.
Thus tan t =1. t = π/4 + kπ. Domain: cos t ≠0, ±1. For t=π/4+kπ, cos t = ±√2/2, not ±1, okay. Argument positive? sin t cos t = tan t cos^2 t = 1 * cos^2 t = cos^2 t >0. So automatically positive. So all k such that t∈[0,2π]: k=0,1. t=π/4, 5π/4. Then x=t/2 = π/8, 5π/8. Both in [0,π]. So 2 solutions. Option B.
But wait let's verify if any other solutions due to base =1? Base =1 gives log undefined (base cannot be 1). So no. Also argument zero? If argument=0, log undefined. Our equation would not hold. But tan=1 gives argument=base>0.
But maybe there is subtlety: log_{b}(a)=1 could also if a=b and b>0 b≠1. Yes.
Let's test original equation for x=π/8: 4x=π/2. cos(4x)=0 => base=0.5*0+0.5=0.5. sin(4x)=1 => argument=0.5. log_{0.5}(0.5)=1. Yes.
x=5π/8: 4x=5π/2. cos(5π/2)=0 => base=0.5. sin(5π/2)=1 => argument=0.5. log=1. Yes.
Are there solutions where sin(4x) = base but base negative? No base must be positive. base = 0.5(cos4x+1) = cos^2(2x) >=0. To be >0, cos4x ≠ -1. If cos4x=-1, base=0 invalid. So exclude.
What about x such that cos4x =1? Then base=1 invalid, log base 1 undefined. But equation maybe? If base=1, log base 1 is undefined. So no.
Thus answer B.
Need craft final response in Portuguese with explanation. Include steps. End with "Resposta: B". Ensure correct.
Let's produce concise explanation.
A equação é:
\log_{(0,5\cos(4x)+0,5)}(0,5\operatorname{sen}(4x))=1
Simplificando a base:
0,5\cos(4x)+0,5=\frac{\cos(4x)+1}{2}=\cos^2(2x)
e o logaritmando:
0,5\operatorname{sen}(4x)=\frac{\operatorname{sen}(4x)}{2}=\operatorname{sen}(2x)\cos(2x).
Seja t=2x. Como x\in[0,\pi], temos t\in[0,2\pi]. A equação fica:
\log_{\cos^2 t}(\operatorname{sen}t\cos t)=1.
Para o logaritmo existir, precisamos de:
\cos^2 t>0,\quad \cos^2 t\neq 1,\quad \operatorname{sen}t\cos t>0.
A equação \log_b(a)=1 implica a=b. Logo:
\operatorname{sen}t\cos t=\cos^2 t.
Como \cos t\neq 0, dividimos por \cos^2 t:
\tan t=1.
Assim,
t=\frac{\pi}{4}+k\pi.
No intervalo t\in[0,2\pi], temos:
t=\frac{\pi}{4}\quad \text{ou}\quad t=\frac{5\pi}{4}.
Voltando para x:
x=\frac{\pi}{8}\quad \text{ou}\quad x=\frac{5\pi}{8}.
Ambos pertencem a [0,\pi] e satisfazem as condições de existência do logaritmo.
Portanto, há 2 valores de x.
Resposta: B
Respondeu
levou 30 s
Como pensou
**Analyzing Equation Domain**
My current focus is on thoroughly defining the domain for the given logarithmic equation. I'm meticulously scrutinizing the conditions for the base ($0.5 \cdot \cos(4x) + 0.5 > 0$ and $0.5 \cdot \cos(4x) + 0.5 \neq 1$) and the argument ($0.5 \cdot \operatorname{sen}(4x) > 0$). This step is crucial before proceeding to solve for $x$.
**Exploring Equation Equality**
I've discovered a simplification for the base, recognizing $0.5 \cos(4x) + 0.5$ as $\cos^2(2x)$. This leads to equating the logand with the base, yielding $\sin(4x) - \cos(4x) = 1$ or $\cos(2x)[\sin(2x) - \cos(2x)] = 0$. The first case, $\cos(2x) = 0$, is invalid due to base existence constraints.
**Solving for Tangent**
I've confirmed that the case $\cos(2x)=0$ is invalid, leaving $\tan(2x)=1$ as the sole avenue. Further validation reveals that all solutions derived from $\tan(2x)=1$ satisfy the original domain constraints for both the base and the argument.
**Verifying Equation Roots**
I'm confirming that the solutions stemming from $\tan(2x)=1$ indeed fulfill the initial equation's requirements. My current analysis focuses on the case where $\sin(4x) - \cos(4x) = 1$, which I've resolved to $\sin(4x - \frac{\pi}{4}) = \frac{\sqrt{2}}{2}$. This is providing a clearer path to the final solution set.
**Confirming Solution Viability**
I've successfully ruled out the solutions arising from $4x = \frac{3\pi}{4} + 2k\pi$ as they fail the base and argument domain constraints. My current focus is confirming that the remaining solutions, $x = \frac{\pi}{8} + \frac{k\pi}{2}$ for $k \in \mathbb{Z}$, are indeed the only valid ones within the specified interval.
**Finalizing Solution Set**
I've completed the analysis, confirming that only two values of $x$, specifically $\frac{\pi}{8}$ and $\frac{5\pi}{8}$, satisfy the original equation within the interval $[0, \pi]$. This completes my verification process for the problem.
Para resolver a equação:
\log_{(0,5 \cdot \cos(4x) + 0,5)}(0,5 \cdot \operatorname{sen}(4x)) = 1
primeiramente analisamos as condições de existência do logaritmo:
-
A base deve ser positiva e diferente de 1:
0,5 \cdot \cos(4x) + 0,5 > 0 \implies \cos(4x) > -1 \implies 4x \neq \pi + 2k\pi, \; k \in \mathbb{Z}
0,5 \cdot \cos(4x) + 0,5 \neq 1 \implies \cos(4x) \neq 1 \implies 4x \neq 2k\pi, \; k \in \mathbb{Z}
-
O logaritmando deve ser estritamente positivo:
0,5 \cdot \operatorname{sen}(4x) > 0 \implies \operatorname{sen}(4x) > 0
Pela definição de logaritmo, \log_b(a) = 1 \iff a = b:
0,5 \cdot \operatorname{sen}(4x) = 0,5 \cdot \cos(4x) + 0,5
Multiplicando toda a equação por 2:
\operatorname{sen}(4x) - \cos(4x) = 1
Podemos reescrever a equação dividindo ambos os lados por \sqrt{2}:
\frac{\sqrt{2}}{2}\operatorname{sen}(4x) - \frac{\sqrt{2}}{2}\cos(4x) = \frac{\sqrt{2}}{2}
\operatorname{sen}\left(4x - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
Isso gera dois conjuntos de soluções:
- 4x - \frac{\pi}{4} = \frac{\pi}{4} + 2k\pi \implies 4x = \frac{\pi}{2} + 2k\pi
- 4x - \frac{\pi}{4} = \frac{3\pi}{4} + 2k\pi \implies 4x = \pi + 2k\pi
Analisando cada caso:
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Para o caso 2: 4x = \pi + 2k\pi
Temos \cos(4x) = -1 e \operatorname{sen}(4x) = 0, o que tornaria a base e o logaritmando iguais a zero, violando as condições de existência. Portanto, essas soluções são descartadas.
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Para o caso 1: 4x = \frac{\pi}{2} + 2k\pi
Temos \cos(4x) = 0 e \operatorname{sen}(4x) = 1. A base e o logaritmando tornam-se:
\text{base} = 0,5 \cdot 0 + 0,5 = 0,5 \quad (\text{válida, pois } 0,5 > 0 \text{ e } 0,5 \neq 1)
\text{logaritmando} = 0,5 \cdot 1 = 0,5 \quad (\text{válido, pois } 0,5 > 0)
Assim, as soluções válidas são dadas por:
x = \frac{\pi}{8} + \frac{k\pi}{2}, \quad k \in \mathbb{Z}
Buscamos as soluções no intervalo [0, \pi]:
- Para k = 0: x = \frac{\pi}{8} \in [0, \pi]
- Para k = 1: x = \frac{5\pi}{8} \in [0, \pi]
- Para outros valores inteiros de k, x não pertence a [0, \pi].
Portanto, há exatamente 2 valores de x que satisfazem a equação.
Resposta: B
