Solve the equation for x on the interval [0,2pi]

Question

The equation is cos^2(x)-sin^2(x)+sin(x)=0

Answer 1

Use the Pythagorean Identity cos²(x) + sin²(x) = 1

That means cos²(x) = 1 - sin²(x) so plug "1 - sin²(x)" in for "cos²(x)" in your equation:

cos²(x) - sin²(x) + sin(x) = 0

(1 - sin²(x)) - sin²(x) + sin(x) = 0

-2sin²(x) + sin(x) + 1 = 0

2sin²(x) - sin(x) - 1 = 0

(2sin(x) + 1)(sin(x) - 1) = 0

Set each piece equal to zero to get:

2sin(x) + 1 = 0 or sin(x) = -1/2 which occurs at 7π/6 and 11π/6

sin(x) - 1 = 0 or sin(x) = 1 which occurs at π/2

So x = π/2, 7π/6, 11π/6

Answer 2

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