Question

[
2^{sin ^{2} x}+4.2^{cos ^{2} x}=6
]
( 2^{sin ^{2} x}+42^{left(1-sin ^{2} xright)}=6 )
( 2^{sin ^{4} x}+6.2 sin ^{2} x+8=0 )
( 2^{sin ^{4} x}+4 cdot frac{3}{2} sin ^{2} x-2 cdot sin ^{2} x+8=0 )
[
2^{sin ^{2} x}left(2^{sin ^{2} x}-4right)-2left(2^{sin ^{2} x}-4right)=0
]
( left(2^{sin ^{2} x}-2right)left(2^{sin ^{2} x}-4right)=0 )
Hence there will be 4 values of ( x )

# (A) (B) 271 toubt (13) Number of values of 'x' in (-21, 21) satisfying the equation 2 4.2 (D) 2 (C)4 (A) 8 (B) 6

Solution