Question

cos 2θ = 13 sin θ − 6 solve the equation

Answer 1

We have:
cos(2θ) = 1 - 2sin^2θ.

So, the equation becomes:
1 - 2sin^2θ = 13sinθ - 6 ==> 2sin^2θ + 13sinθ - 7 = 0.

The left side factors to (2sinθ - 1)(sinθ + 7), so:
(2sinθ - 1)(sinθ + 7) = 0.

By the zero-product property:
2sinθ - 1 = 0 and sinθ + 7 = 0 ==> sinθ = 1/2 and sinθ = -7.

Since -1 ≤ sinθ ≤ 1 for all real θ, sinθ = -7 has no real solutions. From the unit circle, sinθ = 1/2 solves to yield θ = π/6 + 2πk and θ = 5π/6 + 2πk, where k is an integer.

Therefore, the solutions are θ = π/6 + 2πk and θ = 5π/6 + 2πk, where k is an integer.

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Answer 2

thank you!