Question

find the x and y intercept of sinx-(1/18)sin3x show all work please

Answer 1

\[y = \sin x - (1/18)\sin 3x\]
y intercept is when x = 0, this gives y = sin(0) - (1/18)sin(3*0) = 0
so y intercept is y = 0.

x intercept is when y = 0, this gives:
\[0 = \sin x - (1/18)\sin 3x\]
\[\sin x = (1/18)\sin 3x\]
\[18\sin x = \sin 3x = \sin (2x+x) = \sin 2x\cos x + \sin x\cos 2x\]
\[18\sin x = 2\sin x\cos x\cos x + \sin x (\cos ^{2}x - \sin ^{2}x)\]
\[18\sin x = \sin x(2\cos ^{2}x + \cos ^{2}x - \sin ^{2}x) = \sin x(3\cos ^{2}x - \sin ^{2}x)\]
\[18\sin x = \sin x(3(1 - \sin ^{2}x) - sin ^{2}x) = \sin x(3 - 3\sin ^{2}x - \sin ^{2}x)\]
\[18\sin x = \sin x(3 - 4\sin ^{2}x)\]
\[\sin x(18 - 3 + 4\sin ^{2}x) = 0\]
\[\sin x(15 + 4\sin ^{2}x) = 0\]
\[\sin x = 0\]or \[15 + 4\sin ^{2}x = 0\]
2nd equation has no real solutions, so only solution is \[\sin x = 0\]
which gives x intercepts of \[..., -n\pi,-2\pi,-\pi,0,\pi,2\pi,...,n\pi,...\]