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Question

Mathematics Question on Trigonometric Functions

Let B=2 sin⁡2 x−cos⁡ 2xB = 2 \, \sin^2 \, x - \cos \, 2x, then

A

−1≤B≤3 - 1 \le B \le 3

B

0≤B≤2 0 \le B \le 2

C

−1≤B≤1 - 1 \le B \le 1

D

−2≤B≤2- 2 \le B \le 2

Answer

−1≤B≤3 - 1 \le B \le 3

Explanation

Solution

2sin⁡2x−cos⁡2x=2sin⁡2x−(1−2sin⁡2x)=4sin⁡x−12 \sin^{2} x - \cos2x = 2 \sin^{2}x -\left(1 - 2 \sin^{2}x\right) = 4 \sin x - 1 and 0≤sin⁡2x≤1⇔0≤4sin⁡2x≤40 \le\sin^{2} x \le 1 \Leftrightarrow 0 \le 4 \sin^{2} x \le4 ⇔0−1≤4sin⁡2x−1≤3\Leftrightarrow 0-1 \le 4 \sin^{2} x - 1 \le 3