Trigonometry: Solve the equation for all angles

Not too sure what to do here. Please help me out. Thank you.

Solve the equation for all angles, x, in the interval [0degrees, 360degrees).
Equation: -cos2theta = 3 - 3sintheta

There is no need to use the quadratic formula as the second answerer has suggested. It is clear that you mean -cos(2x). Seeing as you would have used ^2 or ² to imply a square.

-cos(2x) = 3 - 3sin(x)
2sin²(x) - 1 = 3 - sin(x)
2sin²(x) + 3sin(x) - 4 = 0

Let u = sin(x)

2u² + 3u - 4 = 0
4u² + 6u - 8 = 0
(2u + 3/2)² - 41/4 = 0
(2u + 3/2)² = 41/4
2u + 3/2 = ± √(41) / 2
2u = -3/2 ± √(41) / 2
u = (-3 ± √(41)) / 4

sin(x) = (-3 ± √(41)) / 4
x = sin‾¹((-3 ± √(41)) / 4)

sin‾¹((-3 - √(41)) / 4) has no solutions because sin(x) cannot exceed |1|.

Hence,
x = sin‾¹((-3 + √(41)) / 4)
x ≈ 58.3°, 121.7°

I have e-mailed you what I did. In essence it was completing the square. Hence the coefficient in front of the x² needs to be a perfect square in order to complete the square. Seeing how using 4 it worked out nicely I multiplied both sides by 2.

However, then you have a constant to deal with.

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That's where the -41/4 comes from you cannot just add (3/2)² , so -9/4 - 8 = -41/4.

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ty

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Is that cos(2x) or cos²(x)? I can't help you unless you clarify.

If:
cos²(x) = 1 - sin²(x)
Use the quadratic formula to solve for sin(x).

If:
cos(2x) = 1 - 2sin²(x)
Use the quadratic formula to solve for sin(x).

No