# Thread: problem set help

1. ## problem set help

solve for 0 < x < 2PI:
a) 2tan^2(x) - 3tan(x) - 4 = 0
b) sin(2x) + sin(4x) = 0
c) tan(2x) + 2cos(x) = 0

prove the identity:
a) sin(4x) + sin(2x) = 2sin(3x)*cos(x)
b) tan(3x) = (3tan(x) - tan^3(x)) / (1 - 3tan^2(x))

thank you for the help

2. Originally Posted by lazyboy292
s

prove the identity:
a) sin(4x) + sin(2x) = 2sin(3x)*cos(x)
b) tan(3x) = (3tan(x) - tan^3(x)) / (1 - 3tan^2(x))
a) Use $\displaystyle \sin a+\sin b=2\sin\frac{a+b}{2}\cos\frac{a-b}{2}$

b) Use $\displaystyle \tan(a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$

3. Hello, lazyboy292!

Solve for $\displaystyle 0 \leq x \leq 2\pi$

. . $\displaystyle c)\;\;\tan(2x) + 2\cos(x) \:=\: 0$

We have: .$\displaystyle \frac{\sin2x}{\cos2x} + 2\cos x \:=\:0$

. . . . $\displaystyle \frac{2\sin x\cos x}{2\cos^2x-1} + 2\cos x \:=\:0$

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

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

. . . . $\displaystyle 2\cos x\bigg(\sin x + 2[1-\sin^2\!x] - 1\bigg)\:=\:0$

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

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

And we have three equations to solve . . .

. . $\displaystyle -2\cos x \:=\:0\quad\Rightarrow\quad \cos x \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}$

. . $\displaystyle \sin x - 1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{2}}$

. . $\displaystyle 2\sin x + 1\:=\:0\quad\Rightarrow\quad \sin x \:=\:-\tfrac{1}{2}\quad\Rightarrow\quad\boxed{ x \:=\:\frac{7\pi}{6},\:\frac{11\pi}{6}}$