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Question

If x,y[0,2π], then find the total number of ordered pairs (x,y) satisfying the equation sinxcosy=1

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Solution

sinxcosy=1
Possible ways are sinx=1, cosy=1 or sinx=1, cosy=1
If sinx=1, cosy=1; hence, x=π/2, y=0,2π
If sinx=1, cosy=1; hence, x=3π/2, y=π
Thus, the possible ordered pairs are (π2,0), (π2,2π) and (3π2,π)
Total number of ordered pairs is 3

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