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Question

Number of solution(s) of the equation 2cos2θ32cosθ+2=0 in (0,10) is

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Solution

2cos2θ32cosθ+2=0
(2cosθ2)(cosθ2)=0
cosθ=12 or cosθ=2 (rejected)
cosθ=12
If θ(0,10), then θ={π4,7π4,9π4}
Number of solutions is 3.

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