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 :)

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- Jan 8th 2013, 08:59 AMkinhew93Trigonometric 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 :) - Jan 8th 2013, 09:36 AMMarkFLRe: Trigonometric equation
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? - Jan 8th 2013, 10:05 AMkinhew93Re: Trigonometric equation
- Jan 8th 2013, 10:17 AMMarkFLRe: Trigonometric equation
Yes, when cos(x) = 0 then sin(x) = ±1 and factors in the form we have cannot be zero.

- Jan 8th 2013, 11:48 AMkinhew93Re: Trigonometric equation
- Jan 9th 2013, 03:28 AMibduttRe: Trigonometric equation
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.