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Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree.

4cos2θ+9=−14cosθ

Find All Angles 0θlt360 That Satisfy The Equation Below To The Nearest 10th Of A Degree 4cos2θ914cosθ class=

Sagot :

abidemiokin

Since cosine is negative in the second and 3rd quadrant, the required angles are 120 and 240 degrees

Trigonometry identity

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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