What is solution 0<x<360 of equation cos2x=2(sinx)^2?

Expert Answers

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You need to use the following double angle identity, such that:

`cos 2x = 1 - 2sin^2 x`

Replacing `1 - 2sin^2 x` for `cos 2x` in the given equation, yields:

`1 - 2sin^2 x = 2sin^2x`

You need to move the terms that contain `sin^2 x` to one side, such that:

`1 = 4sin^2 x => sin^2 x = 1/4 => sin x = +-1/2`

`sin x = 1/2 => {(x = pi/6),(x = pi - pi/6):} =>{(x = pi/6),(x = (5pi)/6):} `

`sin x = -1/2 => {(x = pi + pi/6),(x = 2pi - pi/6):} =>{(x = (7pi)/6),(x = (11pi)/6):} `

Hence, evaluating the solutions to the given equation, under the given conditions, yields `x = pi/6, x = (5pi)/6, x = (7pi)/6, x = (11pi)/6.`

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