Question

Solve $3\cos x+4\sin x=5$, then $x=2n\pi +2awherea={\tan }^{-1}\left(\frac{1}{2}\right),n\in I$
If true then enter $1$ and if false then enter $0$

Answer 1

$\because 3\cos x+4\sin =5$

$\because \cos x=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}$ & $\sin x=\frac{2\tan \frac{x}{2}}{1+{\tan }^3\frac{x}{2}}$

$\therefore$ equation (i) becomes

$\Rightarrow 3\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+4\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)=5$.

Let $\tan \frac{x}{2}=t$

$\therefore$ equation (ii) becomes

$3\left(\frac{1-t^2}{1+t^2}\right)+4\left(\frac{2t}{1+t^2}\right)=5$

$\Rightarrow 4t^2-4t+1=0$

$\Rightarrow {(2t-1)}^2=0$

$\Rightarrow t=\frac{1}{2}\because t=\tan \frac{x}{2}\Rightarrow \tan \frac{x}{2}=\frac{1}{2}\Rightarrow \tan \frac{x}{2}=\tan a,where\tan a=\frac{1}{2}\Rightarrow \frac{x}{2}=n\pi +a$

$\Rightarrow x=2n\pi +2awherea={\tan }^{-1}\left(\frac{1}{2}\right),n\in I$