" 2."sin^(2)x

If 2sin^(2)x + sin^(2)2x =2 then x=

Solve 2sin^(2)x+sin^(2)2x=2

Solve 2sin^(2)x+sin^(2)2x=2 .

Solve: 2sin^(2)x+sin^(2)2x=2

Number of solutions of equation 2"sin"x/2cos^(2)x-2"sin"x/2sin^(2)x=cos^(2)x-sin^(2)x for x in [0,4pi]

" If determinant "|[cos^(2)x,sin^(2)x,cos^(2)x],[sin^(2)x,cos^(2)x,sin^(2)x],[cos^(2)x,sin^(2)x,-cos^(2)x]|" is expanded as a function of "sin^(2)x" ,then the absolute value of constant term in expansion of function "

lim_(x rarr pi/6) (2sin^(2)x + sin x-1)/(2sin^(2)x-3sinx -1) =

3^(sin2x + 2cos^(2)x) + 3^(1 - sin2x + 2sin^(2)x) = 28 is satisfied by :