math help please!!!

Question

solve sin3x + sinx = cosx for 0≤x≤2Pi

Answer 1

sin3x + sinx = cos x
sinx + sin(x + 2x) = cos x
sinx + sinx cos2x + sin2x cos x = cosx
sinx (cos2x + 1) + 2sinx cosx . cosx = cos x  

           

(since 2 cos²x = cos2x +1    and sin 2x = 2sinx cos x)

sinx (2 cos²x) + 2sinxcos²x = cos x
2 sinx cos²x + 2sinxcos²x - cosx = 0
4 sinxcos²x - cosx =0
cosx(4 sinxcosx -1) =0
cosx[2 (2sinxcosx - 0.5)] =0
cosx [2 (sin2x - 0.5)] = 0

Hence cos x = 0 ⇒ x = π/2, 3π/2
or sin2x =0.5 ⇒ 2x = sin-1(0.5) + 2nπ
sin-1 (0.5) = π/6, 5π/6
x = [sin^-1(0.5)]/2 + nπ
When n=0, x = π/12, 5π/12
When n=1, x = 13π/12, 17π/12

Hence, the solution set is {π/12, 5π/12, π/2, 13π/12, 17π/12, 3π/2}

Answer 2

Hi Susan,

 

Is this 3 a multiplier of x or a power?