7sinx=3cos2x?

Question

please help me to solve this

Answer 1

i'm not sure if you mean cos^2(x) or cos(2x)

i'm gonna assume you meant cos(2x)

7sin(x) = 3 cos(2x)

double angle fomula:
cos(2x) = 1 - 2sin^2(x)

substitute
7sin(x) = 3 (1 - 2sin^2(x))

distribute
7sin(x) = 3 - 6sin^2(x)

set the equation equal to 0
6sin^2(x) + 7sin(x) - 3 = 0

let y = sin(x)

6y^2 + 7y - 3 = 0

factor
(2y + 3) (3y - 1) = 0

y = -3/2 or 1/3

recall that y = sin(x)

sin(x) = -3/2
sin(x) = no root because it's less than -1

sin(x) = 1/3
x =~ 0.3398 or 2.80175 <== answer

Answer 2

7sinx=3cos2x
7sin x = 3 (1-2sin^2 x) = 3-6sin^2 x
6sin^2x +7sinx -3 = 0
sin x = [-7 +/- sqrt(7^2-4(6)(-3))]/12
sin x = [-7 +/- sqrt(121)]/12
x = arcsin [-7 +11)]/12 =arcsin(1/3) approx = 19.47 degrees
x also = 180 -19.47 = 160.53 degrees