# Thread: 5 trigo problem [general solution]

1. ## 5 trigo problem [general solution]

p213 ex10a
sorry for my laziness and bad hand writing. thanks

2. $sin(4x)+sin(2x)=0$

$2sin(2x)cos(2x)+sin(2x)=0$

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

$sin(2x)=0$
$2x=0,\pi,2\pi, ....$
$x=0,\frac{\pi}{2},\pi, .... \rightarrow \frac{n\pi}{2}$

$(2cos(2x)+1)=0$
$cos(2x)=\frac{-1}{2}$
$2x=\frac{2\pi}{3},\frac{4\pi}{3},\frac{8\pi}{3},\f rac{10\pi}{3}, ....$
$x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\fra c{5\pi}{3}, ....$
$x=\frac{n\pi}{3}$

3. 18. $\displaystyle\frac{(2n+1)\pi}{3}-\frac{5\pi}{36}=\frac{2n\pi}{3}+\frac{\pi}{3}-\frac{5\pi}{36}=\frac{2n\pi}{3}+\frac{7\pi}{36}$.

4. 19. $(1+\sqrt{3})\sin\theta-(1-\sqrt{3})\cos\theta=2\sqrt{2}$
Multiply both members by $\displaystyle\frac{\sqrt{2}}{4}$.

$\displaystyle\frac{\sqrt{2}+\sqrt{6}}{4}\sin\theta-\frac{\sqrt{2}-\sqrt{6}}{4}\cos\theta=1$

$\displaystyle\sin\frac{7\pi}{12}\sin\theta-\cos\frac{7\pi}{12}\cos\theta=-1$

$\displaystyle\cos\left(\theta+\frac{7\pi}{12}\righ t)=1$

$\displaystyle\theta+\frac{7\pi}{12}=(2n+1)\pi$

$\displaystyle\theta=(2n+1)\pi-\frac{7\pi}{12}=2n\pi+\frac{5\pi}{12}$

5. 27. $\displaystyle\sin x\cos x\cos 2x=\frac{1}{8}\Leftrightarrow8\sin x\cos x\cos 2x=1\Leftrightarrow 4\sin 2x\cos 2x=1\Leftrightarrow 2\sin 4x=1$

Then $\displaystyle\sin 4x=\frac{1}{2}\Rightarrow 4x=(-1)^n\frac{\pi}{6}+n\pi\Rightarrow x=(-1)^n\frac{\pi}{24}+\frac{n\pi}{4}$