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4 years ago

Solve 2cos^2x+sinx-1=0, if 0<=x<=2pi

Mathematics

1 answer:

Tasya [4]4 years ago

3 0

Answer:

x = π/2, 7π/6, 11π/6

Step-by-step explanation:

2 cos² x + sin x − 1 = 0

Use Pythagorean identity:

2 (1 − sin² x) + sin x − 1 = 0

2 − 2 sin² x + sin x − 1 = 0

-2 sin² x + sin x + 1 = 0

2 sin² x − sin x − 1 = 0

Factor:

(sin x − 1) (2 sin x + 1) = 0

Solve:

sin x = 1 or -1/2

x = π/2, 7π/6, 11π/6

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Answer:

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Step-by-step explanation:

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Solve 2cos^2x+sinx-1=0, if 0<=x<=2pi​

Solve 2cos^2x+sinx-1=0, if 0<=x<=2pi