Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree.
4cos2θ+9=−14cosθ

Since cosine is negative in the second and 3rd quadrant, the required angles are 120 and 240 degrees
Trigonometry identities are expressed as a function of cosine, sine and tangent.
Given the trigonometry expression shown
4cos2θ+9=−14cosθ
Equate to zero
4cos2θ+9 + 14cosθ = 0
According to trig identity
cos2θ = 2cos²θ - 1
Substitute to have:
4(2cos²θ - 1)+9 + 14cosθ = 0
Expand
8cos²θ - 4 + 9 + 14cosθ = 0
8cos²θ+ 14cosθ + 5 = 0
let P = cosθ to have;
8P² + 14P + 5 = 0
Factorize the result
8P² + 10P + 4P + 5 = 0
2P(4P+5)+1(4P+5)=0
(2P+1) = 0 and 4P+5 = 0
2P = -1 and P = -5/4
P = -1/2 and -5/4
Recall that P = cosθ
If P = -1/2
cosθ = -1/2
θ = -60
Since cosine is negative in the second and 3rd quadrant, the required angles are 120 and 240 degrees
Learn more on trigonometry identity here: https://brainly.com/question/24349828
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