Trigonometric identities - without using tanx

Question

Question: $\csc x = 4 \sec x$ within the range of $0^\circ-360^\circ$. I attempted to do this without using $\tan x$ - challenging myself.

My attempt: $$\begin{split} \dfrac{1}{\sin x}&= \dfrac{4}{\cos x} \\ 1&= \dfrac{4 \sin x}{\cos x} \\ \cos x &= 4 \sin x \\ 1- \sin x&= 4 \sin x \\ 5 \sin x &= 1 \sin x = 1/5 \\ &\arcsin(1/5) \end{split} $$ This gave me the wrong answer. Is it because $\cos x = 1-\sin x$ is wrong? Is it only possible if sine and cosine are squared? I hope some one can clear this up. Thank you.

Answer 1

\begin{split} &\csc x = \sec x \\ &\dfrac{1}{\sin x} = \dfrac{4}{\cos x} \\ &\cos x -4\sin x = 0 \\ & \sqrt{17}\cos(x+\theta) = 0,\>\>\>\theta = \sin^{-1} \frac{4}{\sqrt{17}}\\ \end{split}

Thus,

$$x= \frac\pi2- \sin^{-1} \frac{4}{\sqrt{17}},\> \frac{3\pi}2 - \sin^{-1} \frac{4}{\sqrt{17}},\>$$

Answer 2

You are correct. The problem is that the identity is $\cos^2 x = 1- \sin^2x$. This problem has no 'nice' solution. Observe $$ \begin{split} \csc x&= 4\sec x \\ \dfrac{1}{\sin x}&= \dfrac{4}{\cos x} \\ \dfrac{1}{4}&= \dfrac{\sin x}{\cos x} \\ \dfrac{1}{4}&= \tan x \\ x&= \tan^{-1}(1/4) \end{split} $$

Answer 3

Without using tan ... $$ \csc x = 4 \sec x \\ \sin x = \frac{\cos x}{4} \\ 16 \sin^2 x = \cos^2 x \\ 16\sin^2 x = 1 - \sin^2 x \\ 17 \sin^2 x = 1 \\ \sin x = \pm\frac{1}{\sqrt{17}} \\ x = \pm \arcsin\frac{1}{\sqrt{17}} $$ Plug in to see that the $+$ is a solution, but the $-$ isn't.