Answer :
To solve the equation \( 2 \sin^2 x - \cos x - 1 = 0 \) on the interval \([0, 2\pi)\), we will use the trigonometric identity \(\sin^2 x = 1 - \cos^2 x\). This will allow us to express the equation in terms of \(\cos x\).
Step-by-Step Solution:
1. Use the identity \(\sin^2 x = 1 - \cos^2 x\):
[tex]\[ 2 \sin^2 x - \cos x - 1 = 0 \implies 2 (1 - \cos^2 x) - \cos x - 1 = 0 \][/tex]
2. Expand and simplify the equation:
[tex]\[ 2 - 2 \cos^2 x - \cos x - 1 = 0 \implies -2 \cos^2 x - \cos x + 1 = 0 \][/tex]
3. Multiply through by -1 to make the quadratic equation more standard:
[tex]\[ 2 \cos^2 x + \cos x - 1 = 0 \][/tex]
4. Solve the quadratic equation for \(\cos x\):
The quadratic equation is \(2y^2 + y - 1 = 0\), where \(y = \cos x\).
To solve for \(y\), we use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -1\).
[tex]\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4} \][/tex]
Thus, we have two solutions:
[tex]\[ y = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
[tex]\[ y = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 \][/tex]
Therefore, \(\cos x = \frac{1}{2}\) or \(\cos x = -1\).
5. Find \(x\) in the interval \([0, 2\pi)\):
- For \(\cos x = \frac{1}{2}\):
[tex]\[ x = \frac{\pi}{3}, \frac{5\pi}{6} \][/tex]
- For \(\cos x = -1\):
[tex]\[ x = \pi \][/tex]
Hence, the complete set of solutions in the interval \([0, 2\pi)\) is:
[tex]\[ x = \frac{\pi}{3}, \frac{5\pi}{3}, \pi \][/tex]
Select the correct choice:
A. \(x = \frac{\pi}{3}, \frac{5\pi}{3}, \pi\)
Therefore, the solutions to the given equation [tex]\(2 \sin^2 x - \cos x - 1 = 0\)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are [tex]\(x = \frac{\pi}{3}, \frac{5\pi}{3}, \pi\)[/tex].
Step-by-Step Solution:
1. Use the identity \(\sin^2 x = 1 - \cos^2 x\):
[tex]\[ 2 \sin^2 x - \cos x - 1 = 0 \implies 2 (1 - \cos^2 x) - \cos x - 1 = 0 \][/tex]
2. Expand and simplify the equation:
[tex]\[ 2 - 2 \cos^2 x - \cos x - 1 = 0 \implies -2 \cos^2 x - \cos x + 1 = 0 \][/tex]
3. Multiply through by -1 to make the quadratic equation more standard:
[tex]\[ 2 \cos^2 x + \cos x - 1 = 0 \][/tex]
4. Solve the quadratic equation for \(\cos x\):
The quadratic equation is \(2y^2 + y - 1 = 0\), where \(y = \cos x\).
To solve for \(y\), we use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -1\).
[tex]\[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4} \][/tex]
Thus, we have two solutions:
[tex]\[ y = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} \][/tex]
[tex]\[ y = \frac{-1 - 3}{4} = \frac{-4}{4} = -1 \][/tex]
Therefore, \(\cos x = \frac{1}{2}\) or \(\cos x = -1\).
5. Find \(x\) in the interval \([0, 2\pi)\):
- For \(\cos x = \frac{1}{2}\):
[tex]\[ x = \frac{\pi}{3}, \frac{5\pi}{6} \][/tex]
- For \(\cos x = -1\):
[tex]\[ x = \pi \][/tex]
Hence, the complete set of solutions in the interval \([0, 2\pi)\) is:
[tex]\[ x = \frac{\pi}{3}, \frac{5\pi}{3}, \pi \][/tex]
Select the correct choice:
A. \(x = \frac{\pi}{3}, \frac{5\pi}{3}, \pi\)
Therefore, the solutions to the given equation [tex]\(2 \sin^2 x - \cos x - 1 = 0\)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are [tex]\(x = \frac{\pi}{3}, \frac{5\pi}{3}, \pi\)[/tex].
