Question

How do you solve `sin(3x) + sin^2 (x)= 2`?

Answer 1

Solve sin 3x + sin^2 x = 2

Ans: `x = (3pi)/2`

Explanation:

`sin 3x + sin^2 x = 2`
Trig identity --> `sin (3x) = 3sin x - 4sin^3 x.` Call sin x = t, we get
`3t - 4t^3 + t^2 = 2`
`f(t) = 4t^3 - t^2 - 3t + 2 = 0`
Since (a - b + c - d = 0), one factor is (t + 1). By division, we get:
`f(t) = (t + 1)(4t^2 - 5t + 2) = 0`. Solve this product.

a. t = sin x = - 1 --> `x = (3pi)/2`
b. `(4t^2 - 5t + 2) = 0`
`D = b^2 - 4ac = 25 - 32 < 0.` There are no real roots.
Therefor, there is unique answer: `x = (3pi)/2`
Check.
`x = (3pi)/2` --> `sin 3x = sin ((9pi)/2) = sin (pi/2) = 1`--> `sin^2 x = 1.`
`sin 3x + sin^2 x = 1 + 1 = 2`. OK