To solve the equation [tex]\(-2 \sin (2x) - 1 = 0\)[/tex] within the interval [tex]\(0 \leq x \leq 2\pi\)[/tex]:
1. Start by isolating the trigonometric function:
[tex]\[-2 \sin (2x) - 1 = 0\][/tex]
Add 1 to both sides of the equation:
[tex]\[-2 \sin (2x) = 1\][/tex]
Divide both sides by -2:
[tex]\[\sin (2x) = -\frac{1}{2}\][/tex]
2. Find the general solutions for [tex]\(2x\)[/tex]:
We need to determine the angles for which the sine function equals [tex]\(-\frac{1}{2}\)[/tex]. The sine function has a value of [tex]\(-\frac{1}{2}\)[/tex] at certain angles.
Recall that [tex]\(\sin(\theta) = -\frac{1}{2}\)[/tex] at:
[tex]\[\theta = \frac{7\pi}{6} + 2k\pi \quad \text{and} \quad \theta = \frac{11\pi}{6} + 2k\pi\][/tex]
for integer values of [tex]\(k\)[/tex].
3. Solve for [tex]\(x\)[/tex]:
Since we have [tex]\(2x = \theta\)[/tex], we'll solve for [tex]\(x\)[/tex]:
[tex]\[
2x = \frac{7\pi}{6} + 2k\pi \implies x = \frac{7\pi}{12} + k\pi
\][/tex]
[tex]\[
2x = \frac{11\pi}{6} + 2k\pi \implies x = \frac{11\pi}{12} + k\pi
\][/tex]
4. Determine the specific solutions within the interval [tex]\(0 \leq x \leq 2\pi\)[/tex]:
- For [tex]\( x = \frac{7\pi}{12} + k\pi \)[/tex]:
- If [tex]\(k = 0\)[/tex]:
[tex]\[
x = \frac{7\pi}{12}
\][/tex]
- If [tex]\(k = 1\)[/tex]:
[tex]\[
x = \frac{7\pi}{12} + \pi = \frac{7\pi}{12} + \frac{12\pi}{12} = \frac{19\pi}{12}
\][/tex]
- For [tex]\( x = \frac{11\pi}{12} + k\pi \)[/tex]:
- If [tex]\(k = 0\)[/tex]:
[tex]\[
x = \frac{11\pi}{12}
\][/tex]
- If [tex]\(k = 1\)[/tex]:
[tex]\[
x = \frac{11\pi}{12} + \pi = \frac{11\pi}{12} + \frac{12\pi}{12} = \frac{23\pi}{12}
\][/tex]
5. List all the solutions:
The solutions within the given interval are:
[tex]\[
x = \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{19\pi}{12}, \frac{23\pi}{12}
\][/tex]
Thus, the values for [tex]\(x\)[/tex] that satisfy the equation [tex]\(-2 \sin (2x) - 1 = 0\)[/tex] within the interval [tex]\(0 \leq x \leq 2\pi\)[/tex] are [tex]\(\frac{7\pi}{12}\)[/tex], [tex]\(\frac{11\pi}{12}\)[/tex], [tex]\(\frac{19\pi}{12}\)[/tex], and [tex]\(\frac{23\pi}{12}\)[/tex].
