Question

Solve the equation $2\sin x+\cos y=2$ for the values of x and y.

• A. $y=2n_1\pi \pm {\cos }^{-1}\left(2\right(1-t\left)\right)$ and $x=n_2\pi +(-1{)}^{n_2}{\sin }^{-1}(t)$
• B. $y=2n_1\pi \pm {\sin }^{-1}(1-t)$ and $x=n_2\pi +(-1{)}^{n_2}{\cos }^{-1}(t)$
• C. $y=n_1\pi \pm {\cos }^{-1}(1-t)$ and $x=2n_2\pi +(-1{)}^{n_2}{\sin }^{-1}(t)$
• D. None of these

Answer 1

The correct option is A $y=2n_1\pi \pm {\cos }^{-1}\left(2\right(1-t\left)\right)$ and $x=n_2\pi +(-1{)}^{n_2}{\sin }^{-1}(t)$

$2\sin x+\cos y=2\Rightarrow 2\sin x=2-\cos y$
$\Rightarrow \sin x=1-\frac{1}{2}\cos y$
$\Rightarrow x=n\pi +{(-1)}^n{\sin }^{-1}(1-\frac{1}{2}\cos y)$
$2\sin x+\cos y=2\Rightarrow \cos y=2-2\sin x\Rightarrow y=2m\pi \pm {\cos }^{-1}(2-2\sin x)$
Where $t=1-\frac{1}{2}\cos y$
$\therefore 2(1-t)=2-2\sin x$
Hence $y=2n_1\pi \pm {\cos }^{-1}\left(2(1-t)\right)x=n_2\pi +{(-1)}^{n_2}{\sin }^{-1}\left(t\right)$