Express 6cos(2x)+sin(x) in terms of sin(x). Hence solve the equation 6cos(2x) + sin(x) = 0, for 0° <= x <= 360°.

Firstly, we need to express 6cos(2x) + sin(x) in terms of sin(x) 6cos(2x) + sin(x) = 6cos(x+x) + sin(x) = 6cos²(x) - 6sin²(x) + sin(x) (applying cos(x+y) = cos(x)cos(y) - sin(x)sin(y)) = 6(1 - sin²(x)) - 6sin²(x) + sin(x) (applying cos²(x) + sin²(x) = 1) = 6 + sin(x) - 12sin²(x)Now to solve 6cos(2x)+sin(x) = 0This is the same as: 6 + sin(x) - 12sin²(x) = 0 12sin²(x) - sin(x) - 6 = 0 (4sin(x) - 3)(3sin(x) + 2) = 0 (Factorising)Therefore sin(x) = 3/4 and sin(x) = -2/3Now, considering the range given is 0° <= x <= 360° : From sin(x) = 3/4, we get x = sin^-1(3/4) = 48.6° And applying the identity sin(x) = sin(180-x) we get an additional solution: 180 - 48.6 = 131.4° The next solution would be 360 + 48.6 = 408.6° but this is outside the range and so we can discard it.From sin(x) = -2/3, we get x = sin^-1(-2/3) = -41.8° This is outside the range, so this will not be our solution, however using the identities for sin we can find solutions within the range. sin(x) = sin(180-x), therefore: -41.8 = 180 - -41.8 = 180 + 41.8 = 221.8° Also, sin(x) = sin(360+x), therefore: -41.8 = 360 - 41.8 = 318.2°And these solutions are our answers: 48.6°, 131.4°, 221.8°, 318.2°