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

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.

#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