Question

How do you solve `1+sinx=2cos^2x`?

Answer 1

`x = 30^o +(-1)^n(180^o xxn)` or `x =-90^o +(360^o xxn)`

where n can be any positive or negative integer, including 0.

Explanation:

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

Substitute `1 - sin^2(x)` for `cos^2(x)`:

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

Distribute the 2:

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

Add `2sin^2(x) - 2` to both sides:

`2sin^2(x) + sin(x) - 1 = 0`

Divide by 2:

`sin^2(x) + 1/2sin(x) - 1/2 = 0`

Use the quadratic formula:

`sin(x) = (-1/2 +-sqrt((1/2)^2 - 4(1)(-1/2)))/(2(1))`

`sin(x) = (-1/2 +-sqrt(1/4 + 2))/(2)`

`sin(x) = (-1/2 +-sqrt(9/4))/(2)`

`sin(x) = (-1/2 +-3/2)/(2)`

`sin(x) = (-1 + 3)/4 and sin(x)=(-1 - 3)/4`

`x = sin^-1(1/2) and sin^-1(-1)`

I am going to use degrees, because the numbers are much nicer than the radian values.

`x = 30^o` or `150^o` and `x=-90^o`

This will repeat an integer multiples of `360^o`

`x = 30^o +(-1)^n(180^o xxn)` or `x =-90^o +(360^o xxn)`

where n can be any positive or negative integer, including 0.

Answer 2

`pi/2 + 2kpi`
`(7pi)/6 + 2kpi`
`(11pi)/6 + 2kpi`

Explanation:

`1 + sin x = 2cos^2 x`
Replace `(2cos^2 x) by (2( 1 - sin^2 x))`:
`1 + sin x = 2( 1 - sin^2 x) = 2 - 2sin^2 x`
Bring quadratic equation to standard form, and solve it
`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`
a. sin x = 1 --> arc `x = pi/2`
b. `sin x = - 1/2`
Trig unit circle gives 2 solutions
`sin x = -1/2` --> arc `x = - pi/6` and arc `x = - (5pi)/6`
Solution for `(0, 2pi)`
`pi/2, (7pi)/6, (11pi)/6`
The arc `(7pi)/6 and (11pi)/6` are co-terminal to the arcs `(- 5pi)/6 and (-pi/6)`.
For general answers, add `2kpi` for each solution