Question
Solve the following equations:
$\cos\text{x}+\cos2\text{x}+\cos3\text{x}=0$
$\cos\text{x}+\cos2\text{x}+\cos3\text{x}=0$
✓
Answer
$\cos\text{x}+\cos2\text{x}+\cos3\text{x}=0$
$\Rightarrow\cos2\text{x}+2\cos2\text{x}.\cos\text{x}=0$ $[\because\cos\text{x}+\cos3\text{x}=2\cos2\text{x}.\cos\text{x}]$
$\Rightarrow\cos2\text{x}(1+2\cos\text{x})=0$
Either
$\cos2\text{x}=0$ or $1+2\cos\text{x}=0$
$\Rightarrow2\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\cos\text{x}=-\frac{1}{2}$
$\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\cos\text{x}=+\cos\Big(\pi-\frac{\pi}{3}\Big)$
or $\cos\text{x}=\cos2\frac{\pi}{3}$
or $\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{z}$
Thus,
$\text{x}=(2\text{n}+1)\frac{\pi}{4}$ or $\Big(2\text{n}\pi\pm\frac{2\pi}{3}\Big),\text{n}\in\text{z}$
$\Rightarrow\cos2\text{x}+2\cos2\text{x}.\cos\text{x}=0$ $[\because\cos\text{x}+\cos3\text{x}=2\cos2\text{x}.\cos\text{x}]$
$\Rightarrow\cos2\text{x}(1+2\cos\text{x})=0$
Either
$\cos2\text{x}=0$ or $1+2\cos\text{x}=0$
$\Rightarrow2\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\cos\text{x}=-\frac{1}{2}$
$\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\cos\text{x}=+\cos\Big(\pi-\frac{\pi}{3}\Big)$
or $\cos\text{x}=\cos2\frac{\pi}{3}$
or $\text{x}=2\text{n}\pi\pm\frac{2\pi}{3},\text{n}\in\text{z}$
Thus,
$\text{x}=(2\text{n}+1)\frac{\pi}{4}$ or $\Big(2\text{n}\pi\pm\frac{2\pi}{3}\Big),\text{n}\in\text{z}$
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