Questão 7

**Explaining simplification options** I'm considering the correct answer, which is B. I see that option A involves kπ and shifted π solutions, which I don't want. I need to explore alternate solving methods to identify x where cos = -1, stating x = 2 arctg2 + 2kπ. It's important to check the original denominator restrictions as well. I feel like I'm on the right track! I'll confirm that Resposta: B is in Portuguese.

Simplifying sin²x/(1−cos x) using the Pythagorean identity gives 1+cos x, which cancels nicely with tan(x/2) written as sin x/(1+cos x), leaving the left side equal to just sin x — so now I need cos x ≠ -1 as a condition and to compare this to the right side, 1 − sin x − cos x. Setting sin x = 1 − sin x − cos x gives 2 sin x + cos x = 1. Substituting t = tan(x/2) using the half-angle formulas turns this into 4t = 2t², so t = 0 or t = 2; t=0 corresponds to cos x = 1, which is excluded, so the valid solution is t = 2, giving x = 2 arctan 2 + 2kπ — matching answer B.

Simplifying sin²x/(1-cosx) reduces nicely to sinx after multiplying by tan(x/2), with domain restrictions cosx≠±1. This turns the equation into 2sinx+cosx=1, which I can tackle using the tangent half-angle substitution t=tan(x/2). Solving the resulting quadratic 2t²-4t=0 gives t=0 or t=2; t=0 corresponds to x=2kπ but that's excluded since cosx=1 there. So the valid solution is t=2, giving x=2arctan2+2kπ, matching answer B.

Simplifying sen²x/(1−cos x) using difference of squares gives 1+cos x, and tg(x/2) equals sen x/(1+cos x), so the left side reduces neatly to sen x. This turns the equation into sen x = 1 − sen x − cos x, which is much simpler to solve. Substituting t = tg(x/2) to convert everything into rational functions of t, the equation simplifies down to 4t = 2t², giving t=0 or t=2. The t=0 case gets excluded since it makes cos x=1, so only t=2 works, giving x=2arctg2+2kπ — matching answer B, and I verify this by plugging back in and confirming both sides equal 4/5.

The problem is from ITA 2027. I need to solve the equation. I identified domain restrictions: x cannot equal 2kπ or π + 2kπ. I will use the Weierstrass substitution t = tan(x/2) to simplify the equation. I simplified the left-hand side using half-angle identities. The fraction \(\frac{\sin^2(x)}{1 - \cos(x)}\) equals \(2\cos^2(x/2)\). This confirms the domain restrictions are consistent. The left-hand side simplifies to \(\sin(x)\) within the domain. I verified this with a numerical check at \(\pi/2\). The equation reduces to 2 sin x + cos x = 1. I will solve this using a sum-to-product identity. I considered rewriting the equation as a single sine function, but the answer options involve 2 arctan expressions. I am now substituting the Weierstrass formulas for sin x and cos x into the equation to solve for t. The Weierstrass substitution yields t = 0 or t = 2. The case t = 0 is extraneous because it falls outside the domain. Only tan(x/2) = 2 remains valid, giving x = 2 arctan(2) + 2kπ for integer k. I verified that the solutions x = 2 arctan(2) + 2kπ satisfy the domain restrictions. The period is 2π, not π. I checked the solution numerically. With tan(x/2)=2, sin x = 0.8 and cos x = -0.6, so both sides of the original equation equal 0.8. This confirms the result. The amplitude of 2 sin x + cos x is √5, which is greater than 1. This means the equation has two solutions per period of 2π. I realized that x = 2kπ satisfies the simplified equation 2 sin x + cos x = 1, but these values are excluded from the original equation's domain because they make 1 - cos x and tan(x/2) zero. This confirms that only x = 2 arctan(2) + 2kπ are valid solutions. I am now checking whether the equation actually has two solutions per period. The case sin(x/2)=0 gives x=2kπ, which is excluded from the domain. The tan(x/2)=2 case yields one solution every 2π. This suggests there may be only one valid solution family, not two. I confirmed that tan(x/2) = 2 has only one solution per period of π, specifically in the interval (0, π/2). This means the equation 2 sin x + cos x = 1 has exactly two solutions per 2π period. I will now solve it using the auxiliary angle method. I solved the equation using the auxiliary angle method. The left-hand side becomes √5 sin(x + α) = 1, where α = arctan(1/2). The general solutions are x + α = arcsin(1/√5) + 2kπ and x + α = π − arcsin(1/√5) + 2kπ. One family gives x = 2kπ, which is excluded from the domain, while the other gives x = 2 arctan(2) + 2kπ. This confirms there are exactly two valid solutions per period of 2π. I am checking whether the algebraic simplification holds when cos(x/2) is negative. For θ = β + π, both sin and cos are negative, so tan remains positive. I need to verify that the identity still works in this case. The algebraic simplification is valid for all values where tan(x/2) is defined. The identity holds even when cos(x/2) is negative, because the cancellation preserves the correct sign. tan(x/2)=2 uniquely fixes sin x and cos x modulo 2π. The Weierstrass formulas confirm that whenever tan(x/2)=2, we have sin x = 4/5 and cos x = -3/5. This means the equation is satisfied for every x in the valid family. I verified that none of the valid solutions equal π + 2kπ or 2kπ, so every member of the family remains defined. The answer is option B.

We need answer question in Portuguese. Need solve. Let's restate equation: Determine all real x satisfying sen^2(x)/(1 - cos x) * tg(x/2) = 1 - sen x - cos x. Options: A x = 2 arctg(2) + kπ, k∈Z B x = 2 arctg(2) + 2kπ C x = 2 arctg(3) + 2kπ D x = 2 arctg(4) + kπ E x = 2 arctg(4) + 2kπ Need choose correct. We need solve carefully domain: denominator 1 - cos x !=0 => cos x !=1 => x != 2kπ. Also tg(x/2) defined => x/2 != π/2 + kπ => x != π + 2kπ? Actually tangent defined except cos(x/2)=0 => x/2 = π/2 + kπ => x = π + 2kπ. So x ≠ π + 2kπ? But tg(x/2) has period 2π? Let's see. Equation: sin^2 x / (1 - cos x) * tan(x/2) = 1 - sin x - cos x. Use identity sin^2 x = 1 - cos^2 x = (1-cos x)(1+cos x). So sin^2 x/(1 - cos x)=1+cos x, provided denominator nonzero. Domain already x≠2kπ. Then LHS = (1+cos x) tan(x/2). Also tan(x/2)= sin x/(1+cos x)? Actually half-angle: tan(x/2)= sin x/(1+cos x) = (1-cos x)/sin x, for cos x ≠ -1? Actually sin x/(1+cos x) if cos x != -1; also equals (1-cos x)/sin x if sin x !=0. Since x≠π+2kπ maybe? Wait tan(x/2) defined if cos(x/2)≠0; but formula sin x/(1+cos x) has denominator 1+cos x; at x=π+2kπ, cos=-1, denominator 0, tan(x/2) undefined? tan(π/2) undefined, yes. But sin x=0 so sin x/(1+cos) 0/0; need exclude. But original tan undefined there, so domain excludes x=π+2kπ. Then LHS = (1+cos x)*(sin x/(1+cos x)) = sin x, for x where 1+cos x≠0, i.e., x≠π+2kπ. Actually equals sin x. Then equation becomes sin x = 1 - sin x - cos x => 2 sin x + cos x = 1? Wait: sin x = 1 - sin x - cos x => 2 sin x + cos x =1. Let's verify: sin x + sin x + cos x =1 => 2 sin x + cos x =1. So need solve 2 sin x + cos x =1 with domain constraints x ≠ 2kπ and x ≠ π+2kπ? Let's solve. 2 sin x + cos x =1. Use t = tan(x/2)? Or linear combination: R = sqrt5, angle φ with cosφ=1/√5? Actually a sin x + b cos x = c. 2 sin x + cos x =1. Write √5 sin(x+α)=1 with tan α? Let's derive. 2 sin x + cos x = R sin(x+θ) where R=√(4+1)=√5, θ = arctan(1/2)? Because sin(x+θ)= sin x cos θ + cos x sin θ. Multiply R: R cosθ sin x + R sinθ cos x =2 sin x +1 cos x => R cosθ=2, R sinθ=1 => tanθ = 1/2, cosθ=2/√5, sinθ=1/√5. So θ = arctan(1/2). Equation √5 sin(x+θ)=1 => sin(x+θ)=1/√5. Then x+θ = arcsin(1/√5) + 2kπ or π - arcsin(1/√5)+2kπ. But arcsin(1/√5)=? θ=arctan(1/2). Note sinθ=1/√5. Since θ∈(0,π/2), arcsin(1/√5)=θ. So solutions: x+θ = θ + 2kπ => x=2kπ (excluded because denominator 1-cos x=0). or x+θ = π - θ + 2kπ => x = π - 2θ + 2kπ = π - 2 arctan(1/2)+2kπ. But π - 2 arctan(1/2) =? Is it 2 arctan(2)? Let's check: arctan(2)+? Since arctan(2)+arctan(1/2)=π/2 (for positive), so 2 arctan(2) + 2 arctan(1/2)=π => π - 2 arctan(1/2)=2 arctan(2). Yes. So x = 2 arctan(2)+2kπ. This is option B. But need check domain: x=2 arctan(2)+2kπ. Does it violate x ≠ π+2kπ? arctan(2)≈63.435°, 2θ≈126.87°, not π? Actually 2 arctan(2)≈126.87° not 180°; plus 2kπ, not π+2kπ? π≈180°, no. Also not 2kπ because 126° not 0 mod 360. So valid. Option B. But let's be careful: Did simplification LHS = sin x valid? Original LHS = sin^2 x/(1-cos x) * tan(x/2). At x where cos x=1, denominator zero, excluded. At x where cos x=-1, sin x=0 denominator 1-cos=2? Wait sin^2 x=0, (1-cos x)=2, quotient 0; tan(x/2)=tan(π/2) undefined. So excluded. At x where sin x=0 but cos x≠1? x=π+2kπ (cos=-1) excluded due tan; x=2kπ excluded due denominator; other multiples? sin x=0 => x=kπ. k even => x=2mπ denominator zero; k odd => x=(2m+1)π = π+2mπ denominator? cos=-1 denominator 2, quotient 0; tan(x/2)=tan(π/2) undefined. So excluded. Fine. But in solving sin x = RHS transformed to 2sin x + cos x =1. Let's verify algebra: Original: (1+cos x) tan(x/2) = 1 - sin x - cos x. tan(x/2)= sin x/(1+cos x) for x ≠ π+2kπ. Then LHS=sin x. Equation sin x = 1 - sin x - cos x => 2 sin x + cos x =1. Correct. But wait, because tan(x/2) formula sin x/(1+cos x) maybe also valid when sin x=0? Let's see at x=π+2kπ: tan(x/2)=tan(π/2 + kπ) undefined; formula gives sin x/(1+cos x)=0/0? Actually 0/0 undefined, so not valid. But excluded. At x=0? formula 0/2=0 but tan(0)=0; okay valid. At x=π? excluded. Good. Alternatively use t=tan(x/2). Let's solve to ensure no missing. Let t=tan(x/2), domain t defined (x ≠ π+2kπ). sin x = 2t/(1+t^2), cos x = (1 - t^2)/(1 + t^2). Also original sin^2/(1-cos) =? sin^2 = (2t/(1+t^2))^2 =4t^2/(1+t^2)^2. 1-cos = 1 - (1-t^2)/(1+t^2)= (1+t^2 -1 + t^2)/(1+t^2)=2t^2/(1+t^2). Quotient = [4t^2/(1+t^2)^2] / [2t^2/(1+t^2)] = 2/(1+t^2), provided t≠0? t=0 => x=2kπ? Actually t=0 tan=0 => x=2kπ? But t=0 gives sin=0 cos=1 denominator zero, excluded. For t≠0. Then LHS = 2/(1+t^2) * t = 2t/(1+t^2). RHS = 1 - 2t/(1+t^2) - (1-t^2)/(1+t^2) = [1+t^2 -2t -1 + t^2]/(1+t^2)= [2t^2 -2t]/(1+t^2)=2t(t-1)/(1+t^2). Equation: 2t/(1+t^2) = 2t(t-1)/(1+t^2) => 2t = 2t(t-1) => if t≠0 (excluded), divide 2t: 1 = t - 1 => t=2. Then x/2 = arctan(2) + kπ => x = 2 arctan(2) + 2kπ? Wait tangent period π for t variable? Let's see t=tan(x/2)=2. General x/2 = arctan(2)+ kπ => x = 2 arctan(2)+2kπ. Yes. Option B. Also t=0 was excluded but check if satisfies original? x=2kπ denominator zero, no. So only B. Need ensure no other solutions from dividing? We divided by t after excluding t=0. But t=0 corresponds x=2kπ, excluded. Good. Option D has kπ not multiple 2; but t=2 gives x/2 = arctan2 + kπ => x = 2 arctan2 + 2kπ, not kπ. If x = 2 arctan2 + kπ? Does any odd multiple satisfy? Let's test x = 2 arctan2 + π. Then x/2 = arctan2 + π/2 => tan(x/2)=tan(arctan2+π/2)= -cot(arctan2)? Actually tan(a+π/2)= -cot a = -1/tan a = -1/2. Then t=-1/2 not 2. Would it satisfy? Let's see original? Maybe not. So only B. Let's verify via direct substitution: x=2 arctan(2). sin x =? sin(2 arctan 2)= 2 tan/(1+tan^2)=4/(1+4)=4/5. cos=(1-tan^2)/(1+tan^2)= (1-4)/5=-3/5. Then LHS: sin^2/(1-cos) = (16/25)/(1+3/5)= (16/25)/(8/5)= (16/25)*(5/8)=80/200=2/5? Wait 16/25*5/8=80/200=2/5=0.4. tan(x/2)=tan(arctan2)=2. LHS=0.4*2=0.8=4/5=sin x. RHS=1 - 4/5 - (-3/5)=1 -0.8+0.6=0.8=4/5. Yes. Good. Options: A +kπ includes π? No, 2 arctan2 + π maybe? Not solution. B +2kπ correct. C arctg3 not. D, E arctg4 not. Answer B. Need provide explanation in Portuguese. Final line "Resposta: B". Include reasoning.

**Solving the Equation** I am currently working through the trigonometric equation involving sine, cosine, and tangent functions. My focus is on simplifying the expression and identifying the real numbers $x$ that satisfy the given condition.