Question

1. $sin(2\theta) = \cos\theta$
2. $cos(2\theta) = 3 - \sin\theta$
3. $cos2x + 6\sin^2x = 4$
4. $cos2x + 5\cos x + 3 = 0$

Explanation:

Problem 7: $\sin(2\theta) = \cos\theta$

Step1: Use double-angle identity

$\sin(2\theta)=2\sin\theta\cos\theta$, substitute:
$2\sin\theta\cos\theta = \cos\theta$

Step2: Rearrange to factor

$2\sin\theta\cos\theta - \cos\theta = 0$
$\cos\theta(2\sin\theta - 1) = 0$

Step3: Solve each factor

1. $\cos\theta = 0$: $\theta = \frac{\pi}{2} + k\pi, k\in\mathbb{Z}$
2. $2\sin\theta - 1 = 0 \implies \sin\theta=\frac{1}{2}$: $\theta = \frac{\pi}{6} + 2k\pi$ or $\theta = \frac{5\pi}{6} + 2k\pi, k\in\mathbb{Z}$

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Problem 8: $\cos(2\theta) = 3 - \sin\theta$

Step1: Use double-angle identity

$\cos(2\theta)=1-2\sin^2\theta$, substitute:
$1-2\sin^2\theta = 3 - \sin\theta$

Step2: Rearrange to quadratic form

$2\sin^2\theta - \sin\theta + 2 = 0$

Step3: Check discriminant

Discriminant $D=(-1)^2-4(2)(2)=1-16=-15<0$
No real solutions.

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Problem 9: $\cos2x + 6\sin^2x = 4$

Step1: Use double-angle identity

$\cos2x=1-2\sin^2x$, substitute:
$1-2\sin^2x + 6\sin^2x = 4$

Step2: Simplify equation

$4\sin^2x + 1 = 4$
$4\sin^2x=3$
$\sin^2x=\frac{3}{4}$
$\sin x=\pm\frac{\sqrt{3}}{2}$

Step3: Solve for $x$

1. $\sin x=\frac{\sqrt{3}}{2}$: $x=\frac{\pi}{3}+2k\pi$ or $x=\frac{2\pi}{3}+2k\pi, k\in\mathbb{Z}$
2. $\sin x=-\frac{\sqrt{3}}{2}$: $x=\frac{4\pi}{3}+2k\pi$ or $x=\frac{5\pi}{3}+2k\pi, k\in\mathbb{Z}$

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Problem 10: $\cos2x + 5\cos x + 3 = 0$

Step1: Use double-angle identity

$\cos2x=2\cos^2x-1$, substitute:
$2\cos^2x-1 + 5\cos x + 3 = 0$

Step2: Simplify to quadratic form

$2\cos^2x + 5\cos x + 2 = 0$

Step3: Factor quadratic

$(2\cos x + 1)(\cos x + 2) = 0$

Step4: Solve each factor

1. $\cos x + 2 = 0 \implies \cos x=-2$: No solution (range of $\cos x$ is $[-1,1]$)
2. $2\cos x + 1 = 0 \implies \cos x=-\frac{1}{2}$: $x=\frac{2\pi}{3}+2k\pi$ or $x=\frac{4\pi}{3}+2k\pi, k\in\mathbb{Z}$

Answer:

1. $\boldsymbol{\theta = \frac{\pi}{2} + k\pi,\ \frac{\pi}{6} + 2k\pi,\ \frac{5\pi}{6} + 2k\pi,\ k\in\mathbb{Z}}$
2. $\boldsymbol{\text{No real solutions}}$
3. $\boldsymbol{x = \frac{\pi}{3} + 2k\pi,\ \frac{2\pi}{3} + 2k\pi,\ \frac{4\pi}{3} + 2k\pi,\ \frac{5\pi}{3} + 2k\pi,\ k\in\mathbb{Z}}$
4. $\boldsymbol{x = \frac{2\pi}{3} + 2k\pi,\ \frac{4\pi}{3} + 2k\pi,\ k\in\mathbb{Z}}$