Answer :
To solve the equation \(\sin(2x) = -\cos(x)\) on the interval \([0, 2\pi)\), we start by using a trigonometric identity for \(\sin(2x)\).
The double-angle identity for sine is:
[tex]\[ \sin(2x) = 2\sin(x)\cos(x) \][/tex]
Given the equation:
[tex]\[ \sin(2x) = -\cos(x) \][/tex]
We substitute the double-angle identity into the equation:
[tex]\[ 2\sin(x)\cos(x) = -\cos(x) \][/tex]
Next, we can simplify this equation. Notice that we have \(\cos(x)\) on both sides of the equation. We can factor out \(\cos(x)\) to further simplify:
[tex]\[ 2\sin(x)\cos(x) + \cos(x) = 0 \][/tex]
[tex]\[ \cos(x)(2\sin(x) + 1) = 0 \][/tex]
For this product to be zero, either \(\cos(x) = 0\) or \(2\sin(x) + 1 = 0\). Let's solve each case separately.
Case 1: \(\cos(x) = 0\)
The cosine function is zero at:
[tex]\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \][/tex]
Case 2: \(2\sin(x) + 1 = 0\)
Solve for \(\sin(x)\):
[tex]\[ 2\sin(x) = -1 \][/tex]
[tex]\[ \sin(x) = -\frac{1}{2} \][/tex]
The sine function is \(-\frac{1}{2}\) at:
[tex]\[ x = \frac{7\pi}{6}, \frac{11\pi}{6} \][/tex]
Combining all solutions within the interval \([0, 2\pi)\), we have the possible values:
[tex]\[ x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} \][/tex]
However, we need to verify these solutions in the original equation \(\sin(2x) = -\cos(x)\). Substituting each back, we find that only \(x = \frac{\pi}{2}\) satisfies the original equation.
Therefore, the solution to the equation \(\sin(2x) = -\cos(x)\) on the interval \([0, 2\pi)\) is:
[tex]\[ \boxed{\frac{\pi}{2}} \][/tex]
The double-angle identity for sine is:
[tex]\[ \sin(2x) = 2\sin(x)\cos(x) \][/tex]
Given the equation:
[tex]\[ \sin(2x) = -\cos(x) \][/tex]
We substitute the double-angle identity into the equation:
[tex]\[ 2\sin(x)\cos(x) = -\cos(x) \][/tex]
Next, we can simplify this equation. Notice that we have \(\cos(x)\) on both sides of the equation. We can factor out \(\cos(x)\) to further simplify:
[tex]\[ 2\sin(x)\cos(x) + \cos(x) = 0 \][/tex]
[tex]\[ \cos(x)(2\sin(x) + 1) = 0 \][/tex]
For this product to be zero, either \(\cos(x) = 0\) or \(2\sin(x) + 1 = 0\). Let's solve each case separately.
Case 1: \(\cos(x) = 0\)
The cosine function is zero at:
[tex]\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \][/tex]
Case 2: \(2\sin(x) + 1 = 0\)
Solve for \(\sin(x)\):
[tex]\[ 2\sin(x) = -1 \][/tex]
[tex]\[ \sin(x) = -\frac{1}{2} \][/tex]
The sine function is \(-\frac{1}{2}\) at:
[tex]\[ x = \frac{7\pi}{6}, \frac{11\pi}{6} \][/tex]
Combining all solutions within the interval \([0, 2\pi)\), we have the possible values:
[tex]\[ x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} \][/tex]
However, we need to verify these solutions in the original equation \(\sin(2x) = -\cos(x)\). Substituting each back, we find that only \(x = \frac{\pi}{2}\) satisfies the original equation.
Therefore, the solution to the equation \(\sin(2x) = -\cos(x)\) on the interval \([0, 2\pi)\) is:
[tex]\[ \boxed{\frac{\pi}{2}} \][/tex]
