We are asked to find all the solutions to the following trigonometric equations: (a) $cos(2\theta) = sin(\theta)$ (b) $2cos(6y) + 11cos(6y)sin(3y) = 0$ (c) $4sin^2(3t) - 3sin(3t) = 1$ Also we need to prove the identity for $tan(\alpha + \beta)$ and find an expression for $tan(2\alpha)$.

1. Problem Description

We are asked to find all the solutions to the following trigonometric equations:

(a) $cos(2\theta) = sin(\theta)$

(b) $2cos(6y) + 11cos(6y)sin(3y) = 0$

(c) $4sin^2(3t) - 3sin(3t) = 1$

Also we need to prove the identity for $tan(\alpha + \beta)$ and find an expression for $tan(2\alpha)$.

2. Solution Steps

(a) $cos(2\theta) = sin(\theta)$

We know that $cos(2\theta) = 1 - 2sin^2(\theta)$.

Substituting, we have $1 - 2sin^2(\theta) = sin(\theta)$.

Rearranging, $2sin^2(\theta) + sin(\theta) - 1 = 0$.

Let $x = sin(\theta)$. Then $2x^2 + x - 1 = 0$.

Factoring, $(2x - 1)(x + 1) = 0$.

So, $x = \frac{1}{2}$ or $x = -1$.

If $sin(\theta) = \frac{1}{2}$, then $\theta = \frac{\pi}{6} + 2\pi k$ or $\theta = \frac{5\pi}{6} + 2\pi k$, where $k$ is an integer.

If $sin(\theta) = -1$, then $\theta = \frac{3\pi}{2} + 2\pi k$, where $k$ is an integer.

(b) $2cos(6y) + 11cos(6y)sin(3y) = 0$

Factoring out $cos(6y)$, we have $cos(6y)(2 + 11sin(3y)) = 0$.

So, either $cos(6y) = 0$ or $2 + 11sin(3y) = 0$.

If $cos(6y) = 0$, then $6y = \frac{\pi}{2} + \pi k$, where $k$ is an integer. Thus, $y = \frac{\pi}{12} + \frac{\pi k}{6}$.

If $2 + 11sin(3y) = 0$, then $sin(3y) = -\frac{2}{11}$.

Thus, $3y = arcsin(-\frac{2}{11}) + 2\pi k$ or $3y = \pi - arcsin(-\frac{2}{11}) + 2\pi k$.

$y = \frac{1}{3}arcsin(-\frac{2}{11}) + \frac{2\pi k}{3}$ or $y = \frac{\pi}{3} - \frac{1}{3}arcsin(-\frac{2}{11}) + \frac{2\pi k}{3}$.

(c) $4sin^2(3t) - 3sin(3t) = 1$

Rearranging, $4sin^2(3t) - 3sin(3t) - 1 = 0$.

Let $x = sin(3t)$. Then $4x^2 - 3x - 1 = 0$.

Factoring, $(4x + 1)(x - 1) = 0$.

So, $x = -\frac{1}{4}$ or $x = 1$.

If $sin(3t) = -\frac{1}{4}$, then $3t = arcsin(-\frac{1}{4}) + 2\pi k$ or $3t = \pi - arcsin(-\frac{1}{4}) + 2\pi k$.

$t = \frac{1}{3}arcsin(-\frac{1}{4}) + \frac{2\pi k}{3}$ or $t = \frac{\pi}{3} - \frac{1}{3}arcsin(-\frac{1}{4}) + \frac{2\pi k}{3}$.

If $sin(3t) = 1$, then $3t = \frac{\pi}{2} + 2\pi k$. Thus, $t = \frac{\pi}{6} + \frac{2\pi k}{3}$.

(d) $tan(\alpha + \beta) = \frac{sin(\alpha + \beta)}{cos(\alpha + \beta)} = \frac{sin\alpha cos\beta + cos\alpha sin\beta}{cos\alpha cos\beta - sin\alpha sin\beta}$.

Dividing the numerator and denominator by $cos\alpha cos\beta$, we get:

$tan(\alpha + \beta) = \frac{\frac{sin\alpha cos\beta}{cos\alpha cos\beta} + \frac{cos\alpha sin\beta}{cos\alpha cos\beta}}{\frac{cos\alpha cos\beta}{cos\alpha cos\beta} - \frac{sin\alpha sin\beta}{cos\alpha cos\beta}} = \frac{tan\alpha + tan\beta}{1 - tan\alpha tan\beta}$.

(e) $tan(2\alpha) = tan(\alpha + \alpha)$.

Using the formula from part (d), we have $tan(2\alpha) = \frac{tan\alpha + tan\alpha}{1 - tan\alpha tan\alpha} = \frac{2tan\alpha}{1 - tan^2\alpha}$.

3. Final Answer

(a) $\theta = \frac{\pi}{6} + 2\pi k$, $\theta = \frac{5\pi}{6} + 2\pi k$, $\theta = \frac{3\pi}{2} + 2\pi k$

(b) $y = \frac{\pi}{12} + \frac{\pi k}{6}$, $y = \frac{1}{3}arcsin(-\frac{2}{11}) + \frac{2\pi k}{3}$, $y = \frac{\pi}{3} - \frac{1}{3}arcsin(-\frac{2}{11}) + \frac{2\pi k}{3}$

(c) $t = \frac{1}{3}arcsin(-\frac{1}{4}) + \frac{2\pi k}{3}$, $t = \frac{\pi}{3} - \frac{1}{3}arcsin(-\frac{1}{4}) + \frac{2\pi k}{3}$, $t = \frac{\pi}{6} + \frac{2\pi k}{3}$

(d) $tan(\alpha + \beta) = \frac{tan\alpha + tan\beta}{1 - tan\alpha tan\beta}$

(e) $tan(2\alpha) = \frac{2tan\alpha}{1 - tan^2\alpha}$