Monoxdifly
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TL;DR Summary: Asking about trigonometric inequality which I'm sure about the result.
The set of all real numbers in the interval [0,2\pi] which satisfy 2sin^2x\geq3cos2x+3 takes the form of [a,b]\cup[c,d]. The value of a + b + c + d + e is ....
A. 4\pi
B. 5\pi
C. 6\pi
D. 7\pi
E. 8\pi
What I've done so far:
2sin^2x\geq3cos2x+3
2sin^2x\geq3(1-2sin^2x)+3
2sin^2x\geq3-6sin^2x)+3
8sin^2x\geq6
sin^2x\geq\frac{3}{4}
How should I continue from this point on?
The set of all real numbers in the interval [0,2\pi] which satisfy 2sin^2x\geq3cos2x+3 takes the form of [a,b]\cup[c,d]. The value of a + b + c + d + e is ....
A. 4\pi
B. 5\pi
C. 6\pi
D. 7\pi
E. 8\pi
What I've done so far:
2sin^2x\geq3cos2x+3
2sin^2x\geq3(1-2sin^2x)+3
2sin^2x\geq3-6sin^2x)+3
8sin^2x\geq6
sin^2x\geq\frac{3}{4}
How should I continue from this point on?