Solve the equation
![4sin ^{2} x + 2sinx - 1 = 0 4sin ^{2} x + 2sinx - 1 = 0](https://tex.z-dn.net/?f=4sin+%5E%7B2%7D+x+%2B+2sinx+-+1+%3D+0)
please solve this problem with explanation.
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Explanation: Call sin x = t --> 4t3+2t2−2t−1= 0 2t2(2t+1)−(2t+1)=0
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