Question

For 0 ≤ ϴ < 2π , what are the solutions to sin2(ϴ) = 2sin2 (StartFraction theta Over 2 EndFraction)?

Answer 1

Answer:

Its A. I took the quiz.

Step-by-step explanation:

Answer 2

Soluions for the given expression in the given interval will be {0, π/2, 3π/2}.

Option A will be the answer.

Solutions of the trigonometric identites:

Given expression in the question,

$\text{sin}^2(\theta)=2\text{sin}^2(\frac{\theta}{2} )$

$[2\text{sin}(\frac{\theta}{2} )\text{cos}(\frac{\theta}{2} )]^2=2\text{sin}^2(\frac{\theta}{2} )$

$4\text{sin}^2(\frac{\theta}{2} )\text{cos}^2(\frac{\theta}{2} )=2\text{sin}^2(\frac{\theta}{2} )$

$4\text{sin}^2(\frac{\theta}{2} )\text{cos}^2(\frac{\theta}{2} )-2\text{sin}^2(\frac{\theta}{2} )=0$

$2\text{sin}^2(\frac{\theta}{2} )[2\text{cos}^2(\frac{\theta}{2} )-1)]=0$

$\text{sin}\frac{\theta}{2}=0$

$\frac{\theta}{2} =n\pi$

$\theta=2n\pi$ [Here, n = integer]

$\theta=0,\pi,2\pi.....$

Therefore, $\theta=0$ [Since, $0\leq \theta < 2\pi$]

$2\text{cos}^2(\frac{\theta}{2})-1=0$

$\text{cos}\theta=0$

$x=\frac{\pi}{2}+\pi n}$ [Here, n = integer]

Therefore, $\theta=\frac{\pi}{2},\frac{3\pi}{2}$ [Since, $0\leq \theta < 2\pi$]

    Hence, soluions for the given expression will be {$0,\frac{\pi}{2} ,\frac{3\pi}{2}$}.

Learn more to get the solutions of the trigonometric expressions here,

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