Oh man I suck at these. Can anyone help out please?
Solve the following equations for 0 ≤ x ≤ 360
a) $\displaystyle sin2x + sinx = 0$
b) $\displaystyle 3cos2x + 2 + cosx = 0$
A kind of method would be fantastic, Thanks guys!
Hello
Remember that $\displaystyle \sin 2x=2 \sin x \cos x$
$\displaystyle 2 \sin x \cos x+ \sin x=0$
Factor by $\displaystyle \sin x$ :
$\displaystyle \sin x (2 \cos x+1)=0$
So either $\displaystyle \sin x=0$, either $\displaystyle 2 \cos x+1=0$
note that $\displaystyle \cos 2x=2 \cos^2 x-1$.b) $\displaystyle 3cos2x + 2 + cosx = 0$
Then substitute $\displaystyle t=\cos x$ and solve for t
We have someone fantastic around hereA kind of method would be fantastic, Thanks guys!