Question

Solve.

2sin^2x = 3 − 3 cosx


Assume that n
is any integer.



x=π+2πn, x=(π/3)+2πn, and x=5π3+2πn
x=(2π/3)+2πn and x=(4π/3)+2πn
x=2πn, x=(π/6)+2πn, and x=(5π/6)+2πn
x=2πn, x=(π/3)+2πn, and x=(5π/3)+2πn
x=(π/3)+2πn, and x=(5π/3)+2πn
There are no solutions to this equation.

Answer 1

To solve the equation \(2\sin^2 x = 3 - 3\cos x\), we first rewrite \(\sin^2 x\) in terms of \(\cos x\) using the Pythagorean identity \(\sin^2 x = 1 - \cos^2 x\). The equation then becomes:

\[
2(1 - \cos^2 x) = 3 - 3\cos x.
\]

Expanding this gives:

\[
2 - 2\cos^2 x = 3 - 3\cos x.
\]

Rearranging the equation leads to:

\[
2\cos^2 x - 3\cos x + 1 = 0.
\]

Now, we can solve this quadratic equation for \(\cos x\). We can use the quadratic formula:

\[
\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]

where \(a = 2\), \(b = -3\), and \(c = 1\). Plugging in these values:

\[
\cos x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2}.
\]

Calculating the discriminant:

\[
\sqrt{9 - 8} = \sqrt{1} = 1.
\]

So we have:

\[
\cos x = \frac{3 \pm 1}{4}.
\]

This gives us two possible solutions:

1. \( \cos x = \frac{4}{4} = 1\)
2. \( \cos x = \frac{2}{4} = \frac{1}{2}\)

Now we solve for \(x\):

1. **For \( \cos x = 1 \)**:
\[
x = 0 + 2\pi n \quad \text{(for any integer } n\text{)}.
\]

2. **For \( \cos x = \frac{1}{2} \)**:
\[
x = \frac{\pi}{3} + 2\pi n \quad \text{or} \quad x = \frac{5\pi}{3} + 2\pi n \quad \text{(for any integer } n\text{)}.
\]

Putting this together, the complete set of solutions is:

\[
x = 2\pi n, \quad x = \frac{\pi}{3} + 2\pi n, \quad x = \frac{5\pi}{3} + 2\pi n.
\]

The corresponding answer from the options provided is:

\[
\boxed{x = 2\pi n, \; x = \frac{\pi}{3} + 2\pi n, \; \text{and} \; x = \frac{5\pi}{3} + 2\pi n.}
\]