1. ## trigo prob

find the most general value of angle A SUCH THAT

3-2cosA-4sinA-cos2A+sin2A=0

2. Hello, prasum!

$\displaystyle \text{Find the most general value of angle }x\text{ such that:}$

.$\displaystyle 3-2\cos x - 4\sin x-\cos2x+\sin2x\:=\:0$

Using Double-angle Identities, we have:

. . . $\displaystyle 3 - 2\cos x - 4\sin x - (1 - 2\sin^2\!x) + 2\sin x\cos x \:=\:0$

. . . . . . . $\displaystyle 2\sin^2\!x - 4\sin x + 2 - 2\cos x + 2\sin x\cos x \;=\;0$

Divide by 2:. . $\displaystyle \sin^2\!x - 2\sin x + 1 + \sin x\cos x - \cos x \;=\;0$

Factor: . . . . . . . . . . . $\displaystyle (\sin x - 1)^2 + \cos x(\sin x - 1) \;=\;0$

Factor:. . . . . . . . . . . . $\displaystyle (\sin x - 1)(\sin x - 1 + \cos x) \;=\;0$

Then we have:

$\displaystyle [1]\;\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2} + 2\pi n$

$\displaystyle [2]\;\sin x -1 + \cos x \:=\:0 \quad\Rightarrow\quad \sin x + \cos x \:=\:1$

Divide by $\displaystyle \sqrt{2}\!:\;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{\sqrt{2}}$

We have: .$\displaystyle \cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x \;=\;\frac{1}{\sqrt{2}}$

. . . . . . . . . . . . . . . . $\displaystyle \sin\left(x + \frac{\pi}{4}\right) \;=\;\frac{1}{\sqrt{2}}$

. . . . . . . . . . . . . . . . . . . .$\displaystyle x + \frac{\pi}{4} \;=\;\begin{Bmatrix}\frac{\pi}{4} + 2\pi n \\ \frac{3\pi}{4} + 2\pi n \end{Bmatrix}$

. . . . . . . . . . . . . . . . . . . . . . .$\displaystyle x \;=\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}$

$\displaystyle \text{Solution: }\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}\;\text{ for any integer }n$

3. Originally Posted by Soroban
Hello, prasum!

Using Double-angle Identities, we have:

. . . $\displaystyle 3 - 2\cos x - 4\sin x - (1 - 2\sin^2\!x) + 2\sin x\cos x \:=\:0$

. . . . . . . $\displaystyle 2\sin^2\!x - 4\sin x + 2 - 2\cos x + 2\sin x\cos x \;=\;0$

Divide by 2:. . $\displaystyle \sin^2\!x - 2\sin x + 1 + \sin x\cos x - \cos x \;=\;0$

Factor: . . . . . . . . . . . $\displaystyle (\sin x - 1)^2 + \cos x(\sin x - 1) \;=\;0$

Factor:. . . . . . . . . . . . $\displaystyle (\sin x - 1)(\sin x - 1 + \cos x) \;=\;0$

Then we have:

$\displaystyle [1]\;\sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:1 \quad\Rightarrow\quad x \:=\:\tfrac{\pi}{2} + 2\pi n$

$\displaystyle [2]\;\sin x -1 + \cos x \:=\:0 \quad\Rightarrow\quad \sin x + \cos x \:=\:1$

Divide by $\displaystyle \sqrt{2}\!:\;\;\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x \;=\;\frac{1}{\sqrt{2}}$

We have: .$\displaystyle \cos\frac{\pi}{4}\sin x + \sin\frac{\pi}{4}\cos x \;=\;\frac{1}{\sqrt{2}}$

. . . . . . . . . . . . . . . . $\displaystyle \sin\left(x + \frac{\pi}{4}\right) \;=\;\frac{1}{\sqrt{2}}$

. . . . . . . . . . . . . . . . . . . .$\displaystyle x + \frac{\pi}{4} \;=\;\begin{Bmatrix}\frac{\pi}{4} + 2\pi n \\ \frac{3\pi}{4} + 2\pi n \end{Bmatrix}$

. . . . . . . . . . . . . . . . . . . . . . .$\displaystyle x \;=\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}$

$\displaystyle \text{Solution: }\;\begin{Bmatrix}2\pi n \\ \frac{\pi}{2} + 2\pi n \end{Bmatrix}\;\text{ for any integer }n$

thanks