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Q.

The number of solutions of the equation sin⁡2θ−2cos⁡θ+4sin⁡θ=4 in [0,5π] is equal to

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a

3

b

4

c

5

d

6

answer is A.

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Detailed Solution

sin⁡2θ−2cos⁡θ+4sin⁡θ=4⇒ (2sin⁡θcos⁡θ+4sin⁡θ)=(2cos⁡θ+4)⇒ sin⁡θ(cos⁡θ+2)=(cos⁡θ+2)∴ sin⁡θ=1⇒ θ=π2,5π2,9π2
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