Solve the equation: sin^2 x + 1 = 2 sin x, for 0 ≤ x < 2π. Sketch the graph of f(x) and g(x).
Understand the Problem
The question appears to be asking for a solution to a trigonometric equation and to sketch the graph of a related function based on the provided graph. It involves solving an equation and demonstrating knowledge of function graphs.
Answer
The solutions are $x = \frac{\pi}{2}, \frac{3\pi}{2}$.
Answer for screen readers
The solutions for the equation $\sin^2 x + 1 = 2 \sin^2 x$ in the interval $[0, 2\pi)$ are:
$$ x = \frac{\pi}{2}, \frac{3\pi}{2} $$
Steps to Solve
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Identify the Trigonometric Equation
Start with the given equation:
$$ \sin^2 x + 1 = 2 \sin^2 x $$
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Simplify the Equation
Rearranging the equation gives:
$$ \sin^2 x + 1 - 2\sin^2 x = 0 $$
This simplifies to:
$$ -\sin^2 x + 1 = 0 $$
Therefore,
$$ \sin^2 x = 1 $$
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Solve for Sine
Taking the square root of both sides:
$$ \sin x = \pm 1 $$
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Find the Values of x
The general solutions for $\sin x = 1$ and $\sin x = -1$ can be expressed as:
- For $\sin x = 1$:
$$ x = \frac{\pi}{2} + 2k\pi $$
- For $\sin x = -1$:
$$ x = \frac{3\pi}{2} + 2k\pi $$
where $k$ is any integer.
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Identify Solutions in the Given Interval
Given that $0 \leq x < 2\pi$, the specific solutions are:
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From $\sin x = 1$: $$ x = \frac{\pi}{2} $$
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From $\sin x = -1$: $$ x = \frac{3\pi}{2} $$
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The solutions for the equation $\sin^2 x + 1 = 2 \sin^2 x$ in the interval $[0, 2\pi)$ are:
$$ x = \frac{\pi}{2}, \frac{3\pi}{2} $$
More Information
The equation deals with the sine function, which oscillates between -1 and 1. The values of $x$ represent the angles at which the sine function reaches its maximum and minimum within a complete cycle of $0$ to $2\pi$.
Tips
- Forgetting to consider both the positive and negative roots of the sine function.
- Not applying the proper general solution format for trigonometric identities.
- Overlooking the specific interval when identifying the final solutions.
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