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Q.

The number of solutions of |cosx|=sinx, such that 4πx4π is:

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a

4

b

6

c

8

d

12

answer is C.

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Detailed Solution

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2 solutions in (0,2π)

So, 8 solutions in [4π,4π]

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The number of solutions of |cos⁡x|=sin⁡x, such that −4π≤x≤4π is: