# Solve equation 2sin^2(3x)+ cos ^2(3x)+sin3x=1

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### 1 Answer

You need to use Pythagorean identity to write the equation in terms of `sin 3x` , such that:

`cos^2 (3x) = 1 - sin^2 (3x)`

Replacing `1 - sin^2 (3x)` for `cos^2 (3x)` yields:

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

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

Reducing duplicate terms yields:

`sin^2 (3x) + sin (3x) = 0`

Factoring out `sin (3x)` yields:

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

Using the zero product rule, yields:

`{(sin (3x) = 0),(sin (3x) + 1 = 0):}`

`3x = (-1)^k*sin^(-1)(0) + n*pi => 3x = n*pi => x = (n*pi)/3`

`sin (3x) + 1 = 0 => sin (3x) = -1 => 3x = (-1)^k*sin^(-1)(-1) + n*pi`

`3x = -pi/2 + n*pi => x = -pi/6 + (n*pi)/3`

**Hence, evaluating the general solutions to the given trigonometric equation yields **`x = -pi/6 + (n*pi)/3, x = (n*pi)/3.`

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