The solution of the trigononometric equation is x = sin⁻¹(-1) = -π/2
How to solve the trigonometric equation?
Since cos²x - sin²x = sinx ; -π < x ≤ π
Solving the trigonometric equation, we have
cos²x - sin²x = sinx
Using sin²x + cos²x = 1 ⇒ cos²x = 1 - sin²x
Substituting this into the equation, we have
cos²x - sin²x = sinx
1 - sin²x - sin²x = sinx
1 - 2sin²x = sinx
1 - 2sin²x - sinx = 0
- 2sin²x - sinx + 1 = 0
Dividing through by -1, we have
- 2sin²x/-1 - sinx/-1 + 1/-1 = 0/-1
2sin²x + sinx - 1 = 0
Let y = sinx
2y² + y - 1 = 0
Factorizing, we have
2y² + 2y - y - 1 = 0
2y(y + 1) - (y + 1) = 0
(2y - 1)(y + 1) = 0
2y - 1 = 0 or y + 1 = 0
2y = 1 or y = -1
y = 1/2 or y = - 1
sinx = 1/2 or sinx = -1
x = sin⁻¹(1/2) = π/6 or x = sin⁻¹(-1) = -π/2
For the interval -π < x ≤ π, the solution is x = sin⁻¹(1/2) = π/6 or x = sin⁻¹(-1) = -π/2
So, the solution of the trigononometric equation is x = sin⁻¹(1/2) = π/6 or x = sin⁻¹(-1) = -π/2
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