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sin 2x-cos x=0

$\sin 2 x-\cos x=0$

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Solve equations involving like terms

Full solution

Q.

\sin 2 x-\cos x=0

Rewrite using double angle formula: Rewrite $\sin 2x$ using the double angle formula: $\sin 2x = 2 \sin x \cos x$ . $\newline$ So the equation becomes $2 \sin x \cos x - \cos x = 0$ .
Factor out $\cos x$ : Factor out $\cos x$ from the equation: $\cos x (2 \sin x - 1) = 0$ .
Set equations equal to zero: Set each factor equal to zero: $\cos x = 0$ and $2 \sin x - 1 = 0$ .
Solve for $\cos x$ : Solve $\cos x = 0$ . The solutions are $x = \frac{\pi}{2} + k\pi$ , where $k$ is an integer.
Solve for $\sin x$ : Solve $2 \sin x - 1 = 0$ . First, add $1$ to both sides: $2 \sin x = 1$ .
Find values of $x$ : Divide both sides by $2$ : $\sin x = \frac{1}{2}$ .
Combine all solutions: Find the values of $x$ that make $\sin x = \frac{1}{2}$ . The solutions are $x = \frac{\pi}{6} + 2k\pi$ and $x = \frac{5\pi}{6} + 2k\pi$ , where $k$ is an integer.
Combine all solutions: Find the values of $x$ that make $\sin x = \frac{1}{2}$ . The solutions are $x = \frac{\pi}{6} + 2k\pi$ and $x = \frac{5\pi}{6} + 2k\pi$ , where $k$ is an integer.Combine all solutions: $x = \frac{\pi}{2} + k\pi$ , $x = \frac{\pi}{6} + 2k\pi$ , and $x = \frac{5\pi}{6} + 2k\pi$ , where $k$ is an integer.

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