# Finding the roots of $\sin x +\cos x - \tan x \sec x$

WolframAlpha tells me that the solutions to $\sin x +\cos x - \tan x \sec x = 0$ are $x = \pi n-\frac{3}{4}\pi, n \in \mathbb Z$, but I cannot see how I would find this myself, other than by plotting the function. Are there identities I can use, or is there even a more general approach to solving trigonometric equations such as this one?

• $\tan x=\sin x/\cos x$, $\sec x=1/\cos x$, $\sin^2x+\cos^x=1$. Aug 31, 2013 at 12:30

$\sin x +\cos x = \tan x \; \sec x \implies \tan x + 1 = \tan x \; \sec^2 x$

Using $t = \tan x$, we have $t + 1 = t(1+t^2) \implies t = 1$

The rest follows...

• Aha, thankyou! And since $\sin^2 x + \cos^2 x = 1$, divide through by $\cos^2 x$ to give $\tan^2 x + 1 = \sec^2 x$, giving $t^2 + 1$ used in the answer. Aug 31, 2013 at 12:44

HINT:

We have $$\sin x+\cos x=\tan x\sec x=\frac{\sin x}{\cos^2x}$$

$$\implies \sin x\cos^2x+\cos^3x=\sin x$$

$$\implies \cos^3x=\sin x-\sin x\cos^2x=\sin x(1-\cos^2x)=\sin^3x$$

$$\implies \tan^3x=1$$

What are the roots of $y^3=1,$ how many of them are real?