# Please solve 3(sinx)^2=cos2x+4sin2x for x.

Asked on by vlk6334

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The equation `3*sin^2 x = cos2x + 4*sin2x` has to be solved for x.

`3*sin^2 x = cos2x + 4*sin2x`

=> `3*sin^2x = cos 2x + 4*sin 2x`

=> 3*(1 - cos 2x)/2 = cos 2x + 4*sin 2x

=> 3*(1 - cos 2x) = 2*cos 2x + 8*sqrt(1 - cos^2 2x)

Let cos 2x = y

=> `3 - 3y = 2y + 8*sqrt(1 - y^2)`

=> `3 - 5y = 8*sqrt(1 - y^2)`

take the square of both the sides

=> `9 + 25y^2 - 30y = 64 - 64y^2`

=> `89y^2 - 30y - 55 = 0`

y1 = `(15 - 32*sqrt 5)/89 `

y2 = `(32*sqrt 5 + 15)/89 `

both the roots lie in [-1, 1]

2x = `cos^-1((15 - 32*sqrt 5)/89)`

=> x = 64.72 degrees

2x = `cos^-1((15 + 32*sqrt 5)/89)`

=> x = 6.73 degrees

It is seen that only for x = 64.72 degrees is the original equation solved.

As the cosine function is periodic in nature the root repeat after every 360 degrees.

The solution of the equation is x = 64.72 + n*360 degrees

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