Question

The value of x and y which satisfies the equation is $4^{sinx}+3^{\frac{1}{cosy}}=11and5.{16}^{sinx}-2.3^{secy}=2$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let, $4^{sinx}=l,3^{\frac{1}{cosy}}=m$
Then the equation becomes

l + m=11 .....(1)
$5l^2-2m=2$ .....(2)
Now apply operation, $2\times \left(1\right)+\left(2\right)$
or, $2l+5l^2=24$
or $5l^2+2l-24=0or5l^2+12l-10l-24=0$
or, $l(5l+12)-2(5l+12)=0$
or, (5l + 12) (l - 2) = 0 So l= 2, -$\frac{12}{5}$ . If l = 2,$4^{sinx}=22^{2sinx}=2;$$\therefore 2sinx=1\therefore sinx=\frac{1}{2}$
If $l=-\frac{12}{5}then{4}^{sinx}=-\frac{12}{5}$ which is impossible for $4^{sinx}<0$
When $l=2$, we get $m=11-2=9$
$\therefore 3^{\frac{1}{cosy}}=3^2$
$\therefore \frac{1}{cosy}=2;cosy=\frac{1}{2}$
When $l=-\frac{12}{5}$, x has no solution. So y has no solution.
Thus we have $sinx=\frac{1}{2},cosy=\frac{1}{2}$
$\therefore x=n\pi +(-1{)}^n\frac{\pi }{6}andy=2m\pi \pm \frac{\pi }{3}$ where m, n $\in$ I.