Question
Solve the following equations:
$2\sin^{2}\text{x}=3\cos\text{x},0\leq\text{x}\leq2\pi$
$2\sin^{2}\text{x}=3\cos\text{x},0\leq\text{x}\leq2\pi$
$\Rightarrow2(1-\cos^{2}\text{x})=3\cos\text{x}$
$\Rightarrow2\cos^{2}\text{x}+3\cos\text{x}-2=0$
$\Rightarrow(2\cos\text{x}-1)(\cos\text{x}+2)=0$
$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-2$
But, $\cos\text{x}=-2$ is not possible. $(-1\leq\cos\text{x}\leq1)$
$\therefore\cos\text{x}=\frac{1}{2}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
Putting n=0 and n=1, We get
$\text{x}=\frac{\pi}{3},\frac{5\text{x}}{3}$ $(0\leq\text{x}\leq2\pi)$
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| Subject | Mathematics | Physics | Chemistry |
| Mean | 42 | 32 | 40.9 |
| Standard | 12 | 15 | 20 |
| Deviation | | |
Using binomial theorem, prove that $3^{2\text{n}+2}-8\text{n}-9$ is divisible by 64 where $\text{n}\in\text{N}.$