Question

Determine the solutions to the equation `2sinx-cos^2x=sin^2x` for `0lexle2pi` ?

Answer 1

`x = pi/6` and `x = (5pi)/6`

Explanation:

Given: `2sin(x)-cos^2(x)=sin^2(x), 0lex < 2pi`

Add `cos^2(x)` to both sides:

`2sin(x)= cos^2(x)+sin^2(x), 0lex < 2pi`

Substitute `1 =cos^2(x)+sin^2(x)`:

`2sin(x)= 1, 0lex < 2pi`

Multiply both sides by `1/2`:

`sin(x) = 1/2, 0lex < 2pi`

Use the inverse sine function on both sides:

`x = sin^-1(1/2), 0lex < 2pi`

The values for this are well known:

`x = pi/6` and `x = (5pi)/6`