1. ## Trigonometric equation

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

2. ## Re: Trigonometric equation

Let's look at the first factor and consider where:

$\sin(\theta)=-\cos(\theta)$

Divide through by $\cos(\theta)$ and so obviously $\cos(\theta)\ne0$:

$\tan(\theta)=-1$

$\theta=\frac{3\pi}{4}+k\pi=\frac{3\pi}{4}(4k+1)$ where $k\in\mathbb{Z}$.

So, what is $\theta$ in this case?

Can you use a similar method for the first factor?

3. ## Re: Trigonometric equation

Originally Posted by MarkFL2

Divide through by $\cos(\theta)$ and so obviously $\cos(\theta)\ne0$:
Do you mean that it is ok to divide by cos because it can't equal 0? (and therefore no solutions will be lost)

4. ## Re: Trigonometric equation

Yes, when cos(x) = 0 then sin(x) = ±1 and factors in the form we have cannot be zero.

5. ## Re: Trigonometric equation

Originally Posted by MarkFL2
Yes, when cos(x) = 0 then sin(x) = ±1 and factors in the form we have cannot be zero.
when cos(x) + sin(x) = 0 I can see how cos(x) cannot equal 0.

But when written as cos(x)(1 + tan(x)) = 0 cos(x) can be 0?

Could someone please try to explain this to me :}

6. ## Re: Trigonometric equation

Cos 5x + sin x = 0
OR
cos 5x + cos ( π/2- x )=0 [cos ( π/2- x )=sin⁡x ]
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.