solve: cos(5x) + sin(x) = 0
I have that:
(sin(3x) + cos(3x))(cos(2x) - sin(2x)) = 0
therefore:
sin(3x) = -cos(3x) and cos(2x) = sin(2x)
from here I'm stuck... any help would be much appreciated
Let's look at the first factor and consider where:
$\displaystyle \sin(\theta)=-\cos(\theta)$
Divide through by $\displaystyle \cos(\theta)$ and so obviously $\displaystyle \cos(\theta)\ne0$:
$\displaystyle \tan(\theta)=-1$
$\displaystyle \theta=\frac{3\pi}{4}+k\pi=\frac{3\pi}{4}(4k+1)$ where $\displaystyle k\in\mathbb{Z}$.
So, what is $\displaystyle \theta$ in this case?
Can you use a similar method for the first factor?
Cos 5x + sin x = 0
OR
cos 5x + cos ( π/2- x )=0 [cos ( π/2- x )=sinx ]
2 cos (5x+ π/2- x )/2 cos (5x-( π/2- x ) )/2=0
[ Because cos C + cos D = 2 cos (C+D)/2 cos (C-D)/2]
That gives
2 cos (4x+ π/2 )/2 cos (6x- π/2 )/2=0
Now from here you can solve the equation.