Solve the equation for exact solutions over the interval [0, 2 $[ 0,2 \pi )$
-$\tan x + \sec x = 1$

Question

Quizplus · Accepted Answer

The equation $\tan x + \sec x = 1$ can be rewritten using the identity $\sec x = \frac{1}{\cos x}$ as $\tan x + \frac{1}{\cos x} = 1$. Recognizing that $\tan x = \frac{\sin x}{\cos x}$, the equation becomes $\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 1$, which simplifies to $\frac{\sin x + 1}{\cos x} = 1$. Multiplying both sides by $\cos x$ gives $\sin x + 1 = \cos x$. Since $\sin x$ and $\cos x$ are equal only at $x = 0$ within the given interval $[0, 2\pi)$, the correct answer is $x = 0$, which corresponds to choice B.

Solve the Equation for Exact Solutions Over the Interval [0 [0,2π)[ 0,2 \pi )[0,2π)

Solve the Equation for Exact Solutions Over the Interval [0 $[ 0,2 \pi )$