Question

How do you solve `cos2A = -sinA - 1`?

Answer 1

Solve f(x) = cos 2x + sin x + 1 = 0

Explanation:

f(x) = (1 - 2sin^2 x) + sin x + 1 = 0. Call sin x = t, we get:
`f(t) = -2t^2 + t + 2 = 0.`
D = d^2 = 1 + 16 = 17 -> d = +- sqrt17# #sin x = t = 1/4 +- (sqrt17)/4 #

a. `sin x = 1/4 - (sqrt17)/4 = -0.78` -->
x = -51.33 or 360 - 51.33 = 308.69 deg
and x = 180 - (-51.33) = 231.33 deg
`b. sin x = 1/4 + (sqrt17)/4` (rejected since > 1)
Answer: Within period (0, 2pi), there are 2 answers: 231.33 and 308.69 deg.
Check.
a. x = 231.33--> cos 2x = cos 462.66 = cos (102.66 + 360) = - 0.22.
--> sin 231.33 = - 0.78.
cos 2x = -0.22 = 0.78 - 1 = - 0.22. OK
b. x = 308.69 --> cos 2x = cos 617.58 = cos (217.58 + 360) = -0.22;
sin 308.69 = -0.78
cos 2x = - sin x - 1
-0.22 = 0.78 - 1 = -0.22 OK.