Question

Find all solutions to
1+cosx=2−2sin^2x
on [0,2π)
.



Select all that apply.


x=4π/3
x=11π/6
x=π/2
x=7π/6
x=0
x=2π/3

Answer 1

To solve the equation \( 1 + \cos x = 2 - 2 \sin^2 x \), we can start by using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \) to express everything in terms of sin or cos.

Since \( \sin^2 x = 1 - \cos^2 x \), we can rewrite \( 2 - 2\sin^2 x \) as follows:

\[
2 - 2\sin^2 x = 2 - 2(1 - \cos^2 x) = 2 - 2 + 2\cos^2 x = 2\cos^2 x
\]

Now, we can rewrite our original equation:

\[
1 + \cos x = 2\cos^2 x
\]

Rearranging gives us a standard quadratic form:

\[
2\cos^2 x - \cos x - 1 = 0
\]

Next, we can solve this quadratic equation for \( \cos x \) using the quadratic formula, \( \cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 2, b = -1, c = -1 \):

\[
\cos x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)}
\]
\[
\cos x = \frac{1 \pm \sqrt{1 + 8}}{4}
\]
\[
\cos x = \frac{1 \pm \sqrt{9}}{4}
\]
\[
\cos x = \frac{1 \pm 3}{4}
\]

This yields two solutions:

\[
\cos x = \frac{4}{4} = 1 \quad \text{and} \quad \cos x = \frac{-2}{4} = -\frac{1}{2}
\]

Now, we can find the values of \( x \):

1. For \( \cos x = 1 \):
\[
x = 0
\]

2. For \( \cos x = -\frac{1}{2} \):
The solutions in the interval \( [0, 2\pi) \) are:
\[
x = \frac{2\pi}{3}, \quad x = \frac{4\pi}{3}
\]

Now we can summarize the solutions to the original equation in the interval \( [0, 2\pi) \):

- \( x = 0 \)
- \( x = \frac{2\pi}{3} \)
- \( x = \frac{4\pi}{3} \)

Now, let's check which of the provided options match these:

1. \( x = 4\pi/3 \) ✔️
2. \( x = 11\pi/6 \) ❌
3. \( x = \pi/2 \) ❌
4. \( x = 7\pi/6 \) ❌
5. \( x = 0 \) ✔️
6. \( x = 2\pi/3 \) ✔️

Thus, the correct answers are:

\[
\text{Solutions: } x = 0, \quad x = \frac{2\pi}{3}, \quad x = \frac{4\pi}{3}
\]

In conclusion, the solutions to the equation \( 1 + \cos x = 2 - 2\sin^2 x \) on the interval \( [0, 2\pi) \) are:

- \( x = 0 \)
- \( x = \frac{2\pi}{3} \)
- \( x = \frac{4\pi}{3} \)