Question

Find angles between 0 and 360 degrees which satisfy the equation sin(x+30 degrees) = 2cosx

Answer 1

270° is not a solution of 2sin2x + cos2x = 0 which can easily be shown by substitution/
x = 270°
2x = 540°
2sin540° + cos540° = 2
2(0) + (-1) = 2
-1 = 2
This is a false statement. Therefore 270° does not satisfy the equation.

Likewise, 45° and 225° are solutions, but 90° is not a solution
Other solutions of the problem as presented are asin(0.6)/2 ≈ 18.4° and 180+asin(0.6)/2 = 198.4°

Verify that you have the problem stated correctly.

Solution of the problem as presented is as follows:.
2sin2x + cos2x = 2
cos2x = 2 - 2sin2x
cos²2x = 4-8sin2x+4sin²2x : square both sides
1-sin²2x = 4-8sin2x + 4sin²2x : apply Pythagorean Identity cos²Θ = 1-sin²Θ
5sin²2x - 8sin2x + 3 = 0
5sin²2x - 5sin2x - 3sin2x + 3 = 0
5sin2x(sin2x-1) - 3(sin2x-1) = 0
(5sin2x - 3)(sin2x-1) = 0

sin2x = 1 → 2x = 90 + 360k → x = 45 + 180k
5sin2x = 3 → 2x = asin(3/5) + 360k → x = asin(3/5)/2 + 180k ≈ 18.4 + 180k

On the interval [0, 360], x = 18.4°, 45°, 198.4°, 225°

Note that when you square both sides of an equation, there may be extraneous solutions to the resulting equation which do not solve the original equation. The solutions obtained in this process must be verified against the original equation. The 4 angle solutions presented above have been verified against and are solutions of the original equation.

Answer 2

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