Averyruss9245
02.12.2019 •
Mathematics
Solve 2cos^2x+sinx-1=0, if 0< =x< =2pi
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Ответ:
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
Ответ:
45
Step-by-step explanation: