and solve $\tan(4x) = \tan(x)$ for $0 < x < 2 \pi$
I think you have to use the addition formulas however I keep on getting stuck on the working so I don't know if I have simply made a mistake or if I'm approaching the question wrong.
Thanks
Best Answer
Consider the first problem:
$$
\sin 4x = \sin x \qquad 0 < x < \pi
$$
Rather than using the addition formula for sine, consider the geometric interpretation of the sine function, being the height of a point on the circumference of the unit circle. This yields
Since $0 < x < \pi$, we need go no further (the next term would be $\sin (x+4\pi)$). So you can now solve
$$
4x = \pi - x
$$
$$
4x = x + 2\pi
$$
$$
4x = 3\pi - x
$$
for three different values of $x$, and those will be your solutions. (Be sure that the domain of solutions really is $0 < x < \pi$, for otherwise $0$ and $\pi$ are also solutions.)
See if you can apply the same kind of thinking for the second problem.
Hint Rebember that
$
\tan( 180^o \pm x ) = \frac{\sin( 180^o \pm x)}{\cos(180^o\pm x)}.
$
Now draw the trigonometric cycle and observes the measures of sine and cosine.
Best Answer
Consider the first problem:
$$ \sin 4x = \sin x \qquad 0 < x < \pi $$
Rather than using the addition formula for sine, consider the geometric interpretation of the sine function, being the height of a point on the circumference of the unit circle. This yields
$$ \sin x = \sin (\pi - x) = \sin (x + 2\pi) = \sin (3\pi-x) = \cdots $$
Since $0 < x < \pi$, we need go no further (the next term would be $\sin (x+4\pi)$). So you can now solve
$$ 4x = \pi - x $$ $$ 4x = x + 2\pi $$ $$ 4x = 3\pi - x $$
for three different values of $x$, and those will be your solutions. (Be sure that the domain of solutions really is $0 < x < \pi$, for otherwise $0$ and $\pi$ are also solutions.)
See if you can apply the same kind of thinking for the second problem.