SOLUTION: solve 4 cos2 θ + 2 sin θ = 3 for 0° ≤ θ ≤ 360°

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Question 1137965: solve 4 cos2 θ + 2 sin θ = 3 for 0° ≤ θ ≤ 360°
Answer by MathLover1(20850) About Me  (Show Source):
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solve 4cos%282theta%29+%2B+2sin%28theta%29+=+3 for 0° ≤ theta ≤ 360°

+4cos%282theta%29+%2B+2sin%28theta%29+-3=0

Use the following identity : cos+%282theta%29=1-2sin+%5E2%28theta%29

4%281-2sin+%5E2%28theta%29%29+%2B+2sin%28theta%29+-3=0

4-8sin+%5E2%28theta%29+%2B+2sin%28theta%29+-3=0

1+%2B+2sin%28theta%29-8sin+%5E2%28theta%29+=0.........let sin+%28theta+%29=u

-8u+%5E2%2B+2u%2B1+=0.....solve with the quadratic formula

u=%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F2a+

u=%28-2%2B-sqrt%282%5E2-4%28-8%29%2A1%29%29%2F%282%28-8%29+%29

u=%28-2%2B-sqrt%284%2B32%29%29%2F%28-16%29

u=%28-2%2B-sqrt%2836%29%29%2F%28-16%29

u=%28-2%2B-6%29%2F%28-16%29

u=%28-1%2B-3%29%2F%28-8%29

solutions:
u=%28-1%2B3%29%2F%28-8%29=>u=2%2F%28-8%29=-1%2F4
u=%28-1-3%29%2F%28-8%29=>u=%28-4%29%2F%28-8%29=1%2F2
so,
sin%28theta+%29=-1%2F4 or sin+%28theta+%29=1%2F2

sin+%28theta+%29=-1%2F4 : if 0%3C=theta+%3C=360° , then

theta=180+-arcsin%281%2F4%29 =>theta=180%2B0.25268=180.25268° ,
and
theta=-+arcsin%281%2F4%29%2B360 ° => theta=+-0.25268+%2B360=359.74732°
sin+%28theta+%29=1%2F2+: if 0%3C=theta+%3C=360° , then theta=30° or theta=150°