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Sagot :
Let's solve the trigonometric equation [tex]\(2 \sin^2(x) - \sin(x) - 1 = 0\)[/tex].
### Step-by-Step Solution:
1. Identify the Type of Equation:
Note that this is a quadratic equation in terms of [tex]\(\sin(x)\)[/tex].
2. Substitute:
Let [tex]\(y = \sin(x)\)[/tex]. Substituting [tex]\(y\)[/tex] into the equation gives:
[tex]\[ 2y^2 - y - 1 = 0 \][/tex]
3. Solve the Quadratic Equation:
Use the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -1\)[/tex].
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{1 + 8}}{4} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{9}}{4} \][/tex]
[tex]\[ y = \frac{1 \pm 3}{4} \][/tex]
This simplifies to two solutions:
[tex]\[ y = \frac{1 + 3}{4} = 1 \][/tex]
[tex]\[ y = \frac{1 - 3}{4} = -\frac{1}{2} \][/tex]
4. Determine the Values of [tex]\(x\)[/tex]:
Recall that [tex]\(y = \sin(x)\)[/tex]. Therefore, we need to solve for [tex]\(x\)[/tex] when [tex]\(\sin(x) = 1\)[/tex] and [tex]\(\sin(x) = -\frac{1}{2}\)[/tex].
- When [tex]\(\sin(x) = 1\)[/tex]:
[tex]\[ x = \frac{\pi}{2} + 2k\pi, \quad \text{for integer } k \][/tex]
Considering the standard interval [tex]\(0 \leq x < 2\pi\)[/tex], we get:
[tex]\[ x = \frac{\pi}{2} \][/tex]
- When [tex]\(\sin(x) = -\frac{1}{2}\)[/tex]:
[tex]\[ x = -\frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{7\pi}{6} + 2k\pi, \quad \text{for integer } k \][/tex]
Considering [tex]\(0 \leq x < 2\pi\)[/tex], we get:
[tex]\[ x = \frac{7\pi}{6}, \quad x = -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6} \quad (\text{but } -\frac{\pi}{6} \text{ is equivalent to } \frac{11\pi}{6} \text{ in the standard range}) \][/tex]
So, the solutions in the standard interval are:
[tex]\[ x = \frac{\pi}{2}, \quad x = \frac{7\pi}{6}, \quad x = \frac{11\pi}{6} \][/tex]
### Final Answer:
The solutions to the equation [tex]\(2 \sin^2(x) - \sin(x) - 1 = 0\)[/tex] are:
[tex]\[ x = -\frac{\pi}{6}, \quad x = \frac{\pi}{2}, \quad x = \frac{7\pi}{6} \][/tex]
### Step-by-Step Solution:
1. Identify the Type of Equation:
Note that this is a quadratic equation in terms of [tex]\(\sin(x)\)[/tex].
2. Substitute:
Let [tex]\(y = \sin(x)\)[/tex]. Substituting [tex]\(y\)[/tex] into the equation gives:
[tex]\[ 2y^2 - y - 1 = 0 \][/tex]
3. Solve the Quadratic Equation:
Use the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 2\)[/tex], [tex]\(b = -1\)[/tex], and [tex]\(c = -1\)[/tex].
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{1 + 8}}{4} \][/tex]
[tex]\[ y = \frac{1 \pm \sqrt{9}}{4} \][/tex]
[tex]\[ y = \frac{1 \pm 3}{4} \][/tex]
This simplifies to two solutions:
[tex]\[ y = \frac{1 + 3}{4} = 1 \][/tex]
[tex]\[ y = \frac{1 - 3}{4} = -\frac{1}{2} \][/tex]
4. Determine the Values of [tex]\(x\)[/tex]:
Recall that [tex]\(y = \sin(x)\)[/tex]. Therefore, we need to solve for [tex]\(x\)[/tex] when [tex]\(\sin(x) = 1\)[/tex] and [tex]\(\sin(x) = -\frac{1}{2}\)[/tex].
- When [tex]\(\sin(x) = 1\)[/tex]:
[tex]\[ x = \frac{\pi}{2} + 2k\pi, \quad \text{for integer } k \][/tex]
Considering the standard interval [tex]\(0 \leq x < 2\pi\)[/tex], we get:
[tex]\[ x = \frac{\pi}{2} \][/tex]
- When [tex]\(\sin(x) = -\frac{1}{2}\)[/tex]:
[tex]\[ x = -\frac{\pi}{6} + 2k\pi \quad \text{and} \quad x = \frac{7\pi}{6} + 2k\pi, \quad \text{for integer } k \][/tex]
Considering [tex]\(0 \leq x < 2\pi\)[/tex], we get:
[tex]\[ x = \frac{7\pi}{6}, \quad x = -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6} \quad (\text{but } -\frac{\pi}{6} \text{ is equivalent to } \frac{11\pi}{6} \text{ in the standard range}) \][/tex]
So, the solutions in the standard interval are:
[tex]\[ x = \frac{\pi}{2}, \quad x = \frac{7\pi}{6}, \quad x = \frac{11\pi}{6} \][/tex]
### Final Answer:
The solutions to the equation [tex]\(2 \sin^2(x) - \sin(x) - 1 = 0\)[/tex] are:
[tex]\[ x = -\frac{\pi}{6}, \quad x = \frac{\pi}{2}, \quad x = \frac{7\pi}{6} \][/tex]
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