Question

Solve `sin x = 2cos^2x - 1` where `x in [0,2pi]`?

Answer 1

`x= pi/6, (5pi)/6, (3pi)/2`

Explanation:

We want to solve:

`sin x = 2cos^2x - 1` where `x in [0,2pi]`

Using the identity `sin^2A+cos^2A = 1` we can write:

`sin x = 2(1-sin^2x) - 1`
`" " = 2-2sin^2x- 1`
`" " = 1-2sin^2x`

`:. 2sin^2x + sinx -1 = 0`

Which is a quadratic in `sinx`, and can be factorised:

`(2sinx -1)(sinx +1) = 0`

Leading to two possible solution:

A) `2sinx -1 => sinx=1/2`
B) `sinx+1 = 0 => sinx=-1`

If we consider the graph of `y=sinx` then we get the following solutions for `x in [0,2pi]`:

Case (A): `\ \ sinx=1/2` has two solutions:

`=> x=pi/6, pi-pi/6`
`" " = pi/6, (5pi)/6`

Case (B): `\ \ sinx=-1` has one solution:

`=> x=(3pi)/2`

Hence, There are three solutions in the specified interval:

`x= pi/6, (5pi)/6, (3pi)/2`

Answer 2

`pi/6; (5pi)/6, (3pi)/2`

Explanation:

`sin x = 2cos^2 x - 1 = cos 2x = 1 - 2sin^2 x`
Bring the equation to standard form:
`2sin^2 x + sin x - 1 = 0`
Since a - b + c = 0, use shortcut. The 2 real roots are:
`sin x = - 1` and `sin x = - c/a = 1/2`
Use trig table and unit circle -->
a. sin x = - 1 --> `x = (3pi)/2`
b. `sin x = 1/2`
`x = pi/6` and `x = (5pi)/6`