A. maths(F.4)

sin^4x + cos^4x

= sin^4x + cos^4x + 2sin²xcos²x - 2sin²xcos²x

= (sin^4x + 2sin²xcos²x + cos^4x) - 2[sinx cosx]²

= (sin²x + cos²x)² - 2[sinx cosx]²

= 1² - 2[sinx cosx]² ............................ as sin²x + cos²x = 1

= 1 - 2[sin(2x)/2]² .............................. as sin(2x) = 2sinx cosx

= 1 - 2[sin²(2x)/4]

= 1 - (1/2)sin²(2x)

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4(sinx^4 + cosx^4) - sin(2x) - 3 = 0

4[1 - (1/2)sin²(2x)] - sin(2x) - 3 = 0 .............. by above part

4 - 2sin²(2x) - sin(2x) - 3 = 0

-2sin²(2x) - sin(2x) + 1 = 0

2sin²(2x) + sin(2x) - 1 = 0

[2sin(2x) - 1][sin(2x) + 1] = 0

2sin(2x) - 1 = 0 or sin(2x) + 1 = 0

sin(2x) = 1/2 or sin(2x) = -1

2x = nπ + (-1)^(n) [π/6] or 2x = nπ + (-1)^(n) [-π/2] for any natural number n

x = nπ/2 + (-1)^(n) [π/12] or x = nπ/2 + (-1)^(n) [-π/4]

x = nπ/2 + (-1)^(n) [π/12] or x = nπ/2 - (-1)^(n) [π/4]

So the general solution is

x = nπ/2 + (-1)^(n) [π/12] or x = nπ/2 - (-1)^(n) [π/4]