solve the equation 2sin x=3x^2+2x+3

2 sin x=3x2+2x+3Since ,-22 sin x2Therefore,-23x2+2x+32-23x2+2x3+123-23x+132+8923-23-89x+13223-89-149x+132-29Square of any real number can't be negative .So there is no real value of x exists.Answer is .

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Range of LHS =[-2,2] RHS To find minima and maxima, differentiate w.r.t x 6x+2=0 x= -1/3 So this will give minimum value of RHS So minimum is 2 So 2sin x=2 sin x=1 x=90° Or else x=n pi
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The answer  is phi
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