Question

Find all solutions to cosx+tanxsinx=−2
on [0,2π).

Enter the answer in terms of π
, as needed. Do not enter a decimal approximation.

Instructions for entering π
.

x=

Answer 1

To solve the equation \( \cos x + \tan x \sin x = -2 \) on the interval \( [0, 2\pi) \), let's first rewrite the equation in terms of sine and cosine.

We know that \( \tan x = \frac{\sin x}{\cos x} \). Thus, we can rewrite the equation as:

\[
\cos x + \frac{\sin^2 x}{\cos x} = -2
\]

Multiplying every term by \( \cos x \) (assuming \( \cos x \neq 0 \)), we get:

\[
\cos^2 x + \sin^2 x = -2 \cos x
\]

Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can replace \( \sin^2 x + \cos^2 x \) with \( 1 \):

\[
1 = -2 \cos x
\]

Now, rearranging gives:

\[
-2 \cos x = 1 \quad \Rightarrow \quad \cos x = -\frac{1}{2}
\]

Now, we need to find \( x \) such that \( \cos x = -\frac{1}{2} \). The cosine function takes this value at:

\[
x = \frac{2\pi}{3}, \quad \frac{4\pi}{3}
\]

Thus, the solutions in the interval \( [0, 2\pi) \) are:

\[
x = \frac{2\pi}{3}, \quad \frac{4\pi}{3}
\]

Therefore, the final answer is:

\[
x = \frac{2\pi}{3}, \frac{4\pi}{3}
\]