If the solutions for \theta from the equation \sin^2 \theta-2

Cameron Russell

Cameron Russell

Answered question

2022-01-26

If the solutions for θ from the equation sin2θ2sinθ+λ=0 lie in nz(2nππ6,(2n+1)π+π6). Then find the possible set values of λ

Answer & Explanation

xleb123

xleb123

Skilled2022-01-28Added 181 answers

Observe that sin(2nππ6)=sin(π6)=sinπ6=? sin((2n+1)π+π6)=sin(π+π6)=sinπ6=? As sinx is increasing near 2nππ6 and decreasing near (2n+1)π+π6, we need sinθ>12sinθ1>32(sinθ1)2<(32)2 Now λ=1(sinθ1)21>1(32)2

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