# Find all the solutions in the interval 0 `<=x<180` cos5x+sin3x-cosx=0

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### 1 Answer

You need to group the terms of equation, such that:

`(cos 5x - cos x) + sin 3x = 0`

You need to convert the difference of cosines into a product, using the following formula, such that:

`cos a - cos b = -2 sin((a+b)/2)sin((a-b)/2)`

Reasoning by analogy, yields:

`cos 5x - cos x = -2sin((5x + x)/2)sin((5x - x)/2)`

`cos 5x - cos x = -2sin 3x*sin 2x`

Hence, substituting `-2sin 3x*sin 2x` for `cos 5x - cos x` yields:

`-2sin 3x*sin 2x + sin 3x = 0`

Factoring out `sin 3x` yields:

`sin 3x(-2sin 2x + 1) = 0`

Since `sin 3x(-2sin 2x + 1) = 0` yields that either `sin 3x = 0` , or `-2sin ` `2x + 1 = 0` , such that:

`sin 3x = 0 => 3x = sin^(-1) 0`

`3x = 0 => x = 0`

`3x = pi ` invalid

`-2sin 2x + 1 = 0 => -2sin 2x = -1 => sin 2x = 1/2`

`2x = sin^(-1)(1/2) => 2x = pi/6 => x = pi/12`

`2x = pi - pi/6 => x = (5pi)/12`

**Hence, evaluating all solutions to equation, in interval `[0,180^o)` , yields `x = 0, x = pi/12, x = (5pi)/12.` **