1. ## Trigonometric equations

Im having a little trouble answering these problems.

Note: Q represents the symbol that looks like a zero with a line across. (Not sure what its called.

1.) 1 - 2 sin(Q) = sin(Q) - sqrt(3)
2.) 2 (sin^2)(Q) = 1 - sin (Q)

2. Originally Posted by ceasar_19134
1.) 1 - 2 sin(Q) = sin(Q) - sqrt(3)
2.) 2 (sin^2)(Q) = 1 - sin (Q)
Call $Q=x$, we have

$1-2\sin x=\sin x-\sqrt3\implies3\sin x=1+\sqrt3$

Can you do it from there?

$2\sin^2x+\sin x-1=0\implies(\sin x+1)(2\sin x-1)=0$

Can you?

3. Originally Posted by ceasar_19134
Note: Q represents the symbol that looks like a zero with a line across. (Not sure what its called.)

1.) 1 - 2 sin(Q) = sin(Q) - sqrt(3)
It's $\theta$ and it's pronounced "theta." We can stick with Q's though.

$1 - 2sin(Q) = sin(Q) - \sqrt{3}$

$-3sin(Q) = -1 - \sqrt{3}$

$sin(Q) = \frac{1}{3} + \frac{\sqrt{3}}{3}$

This has no "nice" solution. So
$Q \approx 1.14494~rad$

-Dan