Trigonometry
Equation
MMTS_Full_Test_05
Grade 12

Question:

The general solution of $\cos 3x\cdot\cos^3 x+\sin 3x\cdot\sin^3 x=0$ is
$(2n+1)\dfrac{\pi}{4}$
$n\pi$
$(2n+1)\dfrac{\pi}{8}$
$\dfrac{n\pi}{4}$

Step-by-Step Solution

Key Concept: $\cos 3x\cos^3 x+\sin 3x\sin^3 x=\cos(3x-x)\cdot$ something; use product-to-sum
$\cos 3x\cos^3 x+\sin 3x\sin^3 x$. Use identities: $=\frac{\cos 2x}{4}(\cos 2x+1)+\frac{\sin 2x}{4}(\ldots)=\frac{1}{4}\cos^2 2x+\frac{\sin 2x\sin 4x}{4}$... Alternate: $=\frac{1}{2}\cos 4x=0\Rightarrow 4x=(2n+1)\pi/2\Rightarrow x=(2n+1)\pi/8$.
Correct Answer: 4

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