Question

How do you solve `1 - 2(sinx)^2 = cosx, 0 <= x <= 360`. Solve for `x`?

Answer 1

`x=0,120,240,360`

Explanation:

`asin^2x+acos^2x-=a`

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

`1-(2-2cos^2x)=cosx`

`1-2+2cos^2x=cosx`

`2cos^2x-cosx-1=0`

Substitute `u=cosx`

`2u^2-u-1=0`

`u=(1+-sqrt((-1)^2-4(2*-1)))/(2*2)`

`u=(1+-sqrt(1-4(-2)))/4`

`u=(1+-sqrt(1+8))/4`

`u=(1+-sqrt(9))/4`

`u=(1+-3)/4`

`u=1or-1/2`

`cosx=1or-1/2`

`x=cos^-1(1)=0,(360-0)=0,360`
`x=cos^-1(-1/2)=120,(360-120)=120,240`

`x=0,120,240,360`

Answer 2

`color(blue)(0, 120^@,240^@,360^@)`

Explanation:

Identity:

`color(red)bb(sin^2x+cos^2x=1)`

Substituting `(1-cos^2x)` in given equation:

`1-2(1-cos^2x)=cosx`

Subtracting `cosx` and expanding:

`1-2+2cos^2x-cosx=0`

Simplify:

`2cos^2x-cosx-1=0`

Let `\ \ \u=cosx`

`:.`

`2u^2-u-1=0`

Factor:

`(2u+1)(u-1)=0=>u=-1/2 and u=1`

But `u=cosx`

`:.`

`cosx=-1/2 , cosx=1`

`x=arccos(cosx)=arccos(-1/2)=>x=120^@`

This is in quadrant II, we also have an angle in quadrant III:

`360^@-120^@=240^@`

`x=arccos(cosx)=arccos(1)=>x=0, 360^@`

Collecting solutions:

`color(blue)(0, 120^@,240^@,360^@)`