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To solve the equation [tex]\( 2 \sin^2 x = 1 - \cos x \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex], let's go through the steps meticulously.
1. Rewrite the equation using a trigonometric identity:
We know that [tex]\(\sin^2 x = 1 - \cos^2 x\)[/tex]. However, it will be helpful to use another identity to simplify our calculations. First, let's rewrite the equation as:
[tex]\[ 2 \sin^2 x + \cos x - 1 = 0. \][/tex]
2. Use the double-angle identity:
Recall that [tex]\( \sin^2 x = \frac{1 - \cos(2x)}{2} \)[/tex]. However, another useful form of transformation is using:
[tex]\[ \sin^2 x = 1 - \cos^2 x. \][/tex]
So, rewriting the expression in terms of [tex]\(\cos x\)[/tex], we have:
[tex]\[ 2(1 - \cos^2 x) + \cos x - 1 = 0. \][/tex]
Simplify this to:
[tex]\[ 2 - 2 \cos^2 x + \cos x - 1 = 0. \][/tex]
Simplify further:
[tex]\[ -2 \cos^2 x + \cos x + 1 = 0. \][/tex]
Notice this is a quadratic equation in terms of [tex]\(\cos x\)[/tex]:
[tex]\[ 2 \cos^2 x - \cos x - 1 = 0. \][/tex]
3. Solve the quadratic equation for [tex]\(\cos x\)[/tex]:
[tex]\[ 2y^2 - y - 1 = 0, \quad \text{where} \quad y = \cos x. \][/tex]
To solve for [tex]\(y\)[/tex], use the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4}. \][/tex]
This gives us two solutions:
[tex]\[ y = \frac{4}{4} = 1 \quad \text{and} \quad y = \frac{-2}{4} = -\frac{1}{2}. \][/tex]
Thus, we have:
[tex]\[ \cos x = 1 \quad \text{and} \quad \cos x = -\frac{1}{2}. \][/tex]
4. Find the corresponding [tex]\(x\)[/tex]-values within the interval [tex]\([0, 2\pi)\)[/tex]:
- For [tex]\(\cos x = 1\)[/tex]:
[tex]\[ x = 0. \][/tex]
- For [tex]\(\cos x = -\frac{1}{2}\)[/tex]:
[tex]\[ x = \frac{2\pi}{3}, \quad \text{and} \quad x = \frac{4\pi}{3}. \][/tex]
5. Gather all the solutions:
Thus, the solutions to the equation [tex]\( 2 \sin^2 x = 1 - \cos x \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}. \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{0, \frac{2\pi}{3}, \frac{4\pi}{3}}. \][/tex]
1. Rewrite the equation using a trigonometric identity:
We know that [tex]\(\sin^2 x = 1 - \cos^2 x\)[/tex]. However, it will be helpful to use another identity to simplify our calculations. First, let's rewrite the equation as:
[tex]\[ 2 \sin^2 x + \cos x - 1 = 0. \][/tex]
2. Use the double-angle identity:
Recall that [tex]\( \sin^2 x = \frac{1 - \cos(2x)}{2} \)[/tex]. However, another useful form of transformation is using:
[tex]\[ \sin^2 x = 1 - \cos^2 x. \][/tex]
So, rewriting the expression in terms of [tex]\(\cos x\)[/tex], we have:
[tex]\[ 2(1 - \cos^2 x) + \cos x - 1 = 0. \][/tex]
Simplify this to:
[tex]\[ 2 - 2 \cos^2 x + \cos x - 1 = 0. \][/tex]
Simplify further:
[tex]\[ -2 \cos^2 x + \cos x + 1 = 0. \][/tex]
Notice this is a quadratic equation in terms of [tex]\(\cos x\)[/tex]:
[tex]\[ 2 \cos^2 x - \cos x - 1 = 0. \][/tex]
3. Solve the quadratic equation for [tex]\(\cos x\)[/tex]:
[tex]\[ 2y^2 - y - 1 = 0, \quad \text{where} \quad y = \cos x. \][/tex]
To solve for [tex]\(y\)[/tex], use the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm \sqrt{1 + 8}}{4} = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4}. \][/tex]
This gives us two solutions:
[tex]\[ y = \frac{4}{4} = 1 \quad \text{and} \quad y = \frac{-2}{4} = -\frac{1}{2}. \][/tex]
Thus, we have:
[tex]\[ \cos x = 1 \quad \text{and} \quad \cos x = -\frac{1}{2}. \][/tex]
4. Find the corresponding [tex]\(x\)[/tex]-values within the interval [tex]\([0, 2\pi)\)[/tex]:
- For [tex]\(\cos x = 1\)[/tex]:
[tex]\[ x = 0. \][/tex]
- For [tex]\(\cos x = -\frac{1}{2}\)[/tex]:
[tex]\[ x = \frac{2\pi}{3}, \quad \text{and} \quad x = \frac{4\pi}{3}. \][/tex]
5. Gather all the solutions:
Thus, the solutions to the equation [tex]\( 2 \sin^2 x = 1 - \cos x \)[/tex] on the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}. \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{0, \frac{2\pi}{3}, \frac{4\pi}{3}}. \][/tex]
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