# Thread: Trigonemetric equation help

1. ## Trigonemetric equation help

??

2. Hello, astrostar

You're almost there . . .

$1 -2\sin^2\!\theta + \cos^2\!\theta \:= \:0\;\text{ for }0^o \leq \theta < 360^o$

My answer so far:

. . $1 + \sin^2\!\theta + \cos^2\!\theta \:= \:3\sin^2\!\theta$

. . $2 \:= \:3\sin^2\!\theta$ . . . . Good!

We have: . $3\sin^2\!\theta \:=\: 2\quad\Rightarrow\quad \sin^2\!\theta \:=\:\frac{2}{3}\quad\Rightarrow\quad \sin\theta \:=\:\pm\sqrt{\frac{2}{3}}$

Therefore: . $\theta \;\;=\;\;\sin^{-1}\left(\pm\sqrt{\frac{2}{3}}\right) \;\;\approx\;\; \left\{54.7^o,\:125.3^o,\:234.7^o,\:305.3^o\right\ }$

3. Originally Posted by astrostar
Hi,

Can you help me finish this equation off. The answer needs to be angles from 0 to 360degrees.

1 - 2 sin^2 E + cos^2 E = 0

My answer so far,

1 + sin^2 E + cos^2 E = 3 sin^2 E

2 = 3 sin^2 E

Any help appreciated
You were on the right track; just continue with what you were doing!

$2=3sin^2(E)$

$sin^2(E)=\frac{2}{3}$

$sin(E)=\pm \sqrt{\frac{2}{3}}$

$E = sinh(\pm \frac{2}{3})$

You'll have to use your calculator for the last part!

4. Thanks for your help