In Exercises 63–84, use an identity to solve each equation on the interval [0, 2𝝅). sin 2x = cos x
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Recall the double-angle identity for sine: \(\sin 2x = 2 \sin x \cos x\). Substitute this into the equation to get \(2 \sin x \cos x = \cos x\).
Rewrite the equation as \(2 \sin x \cos x - \cos x = 0\) to set it equal to zero, which allows factoring.
Factor out \(\cos x\) from the left side: \(\cos x (2 \sin x - 1) = 0\).
Set each factor equal to zero to find possible solutions: \(\cos x = 0\) and \(2 \sin x - 1 = 0\).
Solve each equation separately on the interval \([0, 2\pi)\): For \(\cos x = 0\), find all \(x\) where cosine is zero; for \(2 \sin x - 1 = 0\), solve for \(\sin x = \frac{1}{2}\) and find all corresponding \(x\) values.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that hold true for all values within their domains. In this problem, the double-angle identity for sine, sin(2x) = 2 sin x cos x, is essential to rewrite the equation in a solvable form.
Solving trigonometric equations involves manipulating the equation using identities and algebraic techniques to isolate the variable. After applying identities, solutions are found by setting factors equal to zero or using inverse trigonometric functions within the given interval.
When solving trigonometric equations, solutions must be restricted to the specified interval, here [0, 2π). This means finding all angles within one full rotation of the unit circle that satisfy the equation, considering periodicity and multiple solutions.