1. trig equation

1) cos(40+x)=3sin(50+x) find x for 0<x<360 in degrees

thanks!

2. 1. $\displaystyle cos(a+b) = cos a cos b - sin a sin b$ and $\displaystyle sin(a+b) = sin a cos b + sin b cos a$ so $\displaystyle cos(40+x) = cos 40 cos x - sin 40 sin x$ and $\displaystyle sin(50+x) = sin 50 cos x + cos 50 sin x$. The result is $\displaystyle cos 40 cos x - sin 40 sin x = 3 sin 50 cos x + 3 cos 50 sin x$

$\displaystyle cos 40 cos x - sin 40 sin x = 3 cos 40 cos x + 3 sin 40 sin x$

$\displaystyle 2 cos 40 cos x + 4 sin 40 sin x = 0; 4 sin 40 sin x = -2 cos 40 cos x$

$\displaystyle \frac{-cos 40}{2 sin 40} = \frac{sin x}{cos x}$

$\displaystyle tan x = \frac{-cos 40}{2 sin 40}$

$\displaystyle x = arctan{\frac{-cos 40}{2 sin 40}}$

2. $\displaystyle 8 sin x cos x = 4 sin 2x = 3$

$\displaystyle sin 2x = \frac{3}{4}$

$\displaystyle 2x = arcsin {\frac{3}{4}}$

$\displaystyle x = \frac{1}{2} arcsin {\frac{3}{4}}$

3. please can you explain where did 2cos40cosx and +4sin40sinx come from?

4. Originally Posted by snowball

please can you explain where did 2cos40cosx and +4sin40sinx come from?

$\displaystyle cos 40 cos x - sin 40 sin x = 3 cos 40 cos x + 3 sin 40 sin x$

$\displaystyle -4 sin 40 sin x = 2 cos 40 cos x$

$\displaystyle 0 = 2 cos 40 cos x +4 sin 40 sin x$