Apply sum-to-product identities: Use the sum-to-product identities to simplify the left side of the equation. The sum-to-product identities state that cos A - cos B = -2 sin((A + B)/2) sin((A - B)/2). So, cos 3x - cos 5x can be written as -2 sin((3x + 5x)/2) sin((3x - 5x)/2).Use identity to simplify: Apply the sum-to-product identity to the left side of the equation. cos 3x - cos 5x = -2 sin((3x + 5x)/2) sin((3x - 5x)/2) = -2 sin(4x) sin(-x) Since sin(-x) = -sin(x), this simplifies to: = 2 sin(4x) sin(x)Substitute simplified expression: Substitute the simplified left side back into the original equation. 2 sin(4x) sin(x) = sin 4xDivide by sin(4x): Divide both sides of the equation by sin(4x), assuming sin(4x) is not zero. sin(4x) / sin(4x) = (2 sin(4x) sin(x)) / sin(4x) 1 = 2 sin(x)Isolate sin(x): Divide both sides by 2 to isolate sin(x). 1 / 2 = sin(x)Find general solution: Find the general solution for x. Since sin(x) = 1/2, x can be any angle whose sine is 1/2. This occurs at x = π/6 + 2πn or x = 5π/6 + 2πn, where n is any integer.

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$\cos 3 x-\cos 5 x=\sin 4 x$

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Grade 7

Add and subtract like terms: with exponents

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\cos 3 x-\cos 5 x=\sin 4 x

Apply sum-to-product identities: Use the sum-to-product identities to simplify the left side of the equation. $\newline$ The sum-to-product identities state that $\cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)$ . $\newline$ So, $\cos 3x - \cos 5x$ can be written as $-2 \sin\left(\frac{3x + 5x}{2}\right) \sin\left(\frac{3x - 5x}{2}\right)$ .
Use identity to simplify: Apply the sum-to-product identity to the left side of the equation. $\newline$ $\cos 3x - \cos 5x = -2 \sin\left(\frac{3x + 5x}{2}\right) \sin\left(\frac{3x - 5x}{2}\right)$ $\newline$ $= -2 \sin(4x) \sin(-x)$ $\newline$ Since $\sin(-x) = -\sin(x)$ , this simplifies to: $\newline$ $= 2 \sin(4x) \sin(x)$
Substitute simplified expression: Substitute the simplified left side back into the original equation. $\newline$ $2 \sin(4x) \sin(x) = \sin 4x$
Divide by $\sin(4x)$ : Divide both sides of the equation by $\sin(4x)$ , assuming $\sin(4x)$ is not zero. $\newline$ $\frac{\sin(4x)}{\sin(4x)} = \frac{2 \sin(4x) \sin(x)}{\sin(4x)}$ $\newline$ $1 = 2 \sin(x)$
Isolate $\sin(x)$ : Divide both sides by $2$ to isolate $\sin(x)$ . $\frac{1}{2} = \sin(x)$
Find general solution: Find the general solution for $x$ . $\newline$ Since $\sin(x) = \frac{1}{2}$ , $x$ can be any angle whose sine is $\frac{1}{2}$ . $\newline$ This occurs at $x = \frac{\pi}{6} + 2\pi n$ or $x = \frac{5\pi}{6} + 2\pi n$ , where $n$ is any integer.