Question

Sin^2x+3cos^2x=5sinx

Answer 1

Answer:

sin^2x + 3cos^2x = 5sinx

sin^2x + 3 - 3sin^2x = 5sinx (cos^2x = 1-sin^2x)

3-2sin^2x = 5sinx

2sin^2x + 5sinx - 3 = 0

2sin^2x + 6sinx - sinx - 3= 0

2sinx(sinx + 3) -1(sinx + 3) = 0

(sinx + 3)(2sinx - 1) = 0

this means either

sinx=-3 (this is not possible as value of sinx is between -1 and 1)

2simx -1 =0

sinx = 1/2 = sin π/6

hence general solution is

x = nπ + (-1)^n π/6