Solve for the exact solutions in the interval $[0, 2\pi)$. If the equation has no solutions, respond with DNE.

$\sin(2x) = \cos(x)$

Neetesh Kumar | October 15, 2024

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Assignment 9.5 Question 14: - On Solving Trignometric Equations
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Step-by-step solution:

Step 1: Use the double-angle identity for $\sin(2x)$

Recall that $\sin(2x) = 2\sin(x)\cos(x)$. Substitute this into the equation:

$2\sin(x)\cos(x) = \cos(x)$

Step 2: Rearrange the equation

We can factor out $\cos(x)$:

$\cos(x) (2\sin(x) - 1) = 0$

Step 3: Solve for $x$

This gives us two possible cases:

1. $\cos(x) = 0$
2. $2\sin(x) - 1 = 0$

Case 1: $\cos(x) = 0$

The solutions for $\cos(x) = 0$ in the interval $[0, 2\pi]$ are:

$x = \dfrac{\pi}{2}, \dfrac{3\pi}{2}$

Case 2: $2\sin(x) - 1 = 0$

Solving for $\sin(x)$, we get:

$\sin(x) = \frac{1}{2}$

The solutions for $\sin(x) = \frac{1}{2}$ in the interval $[0, 2\pi]$ are:

$x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}$

Step 4: Final solution

The exact solutions for $x$ are:

$x = \boxed{\dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{\pi}{6}, \dfrac{5\pi}{6}}$

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