Roots of the equation2 sin^2(A) + sin^2(2A) = 2are - (a)p/6 (b)p/4 (c)p/3 (d)p/2 where p is for pi...

hi

in these type of questions it is better to check from the options because there are only two terms 2 sin^2(A) + sin^2(2A)               put A = PIE / 2  you will see that this value satisfies the equation.

and for subjective take the second term sin^2(2A)  = 4 sin^2A * cos^2 A= 4 sin^2A (1-sin^2)...BY TRIGONOMETRIC IDENTITY

so the equation will become

4sin^2(A) - sin^4(A) + 2 sin^2 (A) = 2

put     sin^2A = t    you will get the quadratic equation 

then solve it for sin (A) you will get the answer

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2sin^2(A)+sin^2(2A)=2

2sin^2(A)+4sin^2(A)cos^2(A)=2

2sin^2(A)+4sin^2(A){1-sin^2(A)}=2

sin^2(A)+2sin^2(A){1-sin^2(A)=1

2sin^4(A)-3sin^2(A)+1=0

sin^2(A)=3/4+_(9-8)^1/2/4

sin^2(A)=1/2 or1

sin(A)=+_(1/2)^1/2 Or +_1

A=P/6 OrP/2