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  • The values of x and y satisfying the system of equations  <img alt="{2^{\sin x

The values of x and y satisfying the system of equations  {2^{\sin x + \cos y}} = 1\;,\;{16^{{{\sin }^2}x + {{\cos }^2}y}} = 4 are given by

(i) x = n\pi + {( - 1)^n}\frac{\pi }{6}\;\;{\rm{and}}\;\;y = 2n\pi \pm \frac{\pi }{3}

(ii) x = n\pi + {( - 1)^{n + 1}}\frac{\pi }{6}\;\;{\rm{and}}\;\;y = 2n\pi \pm \frac{\pi }{3}

(iii)x = n\pi + {( - 1)^n}\frac{\pi }{6}\;\;{\rm{and}}\;\;y = 2n\pi \pm \frac{{2\pi }}{3}

(iv)x = n\pi + {( - 1)^{n + 1}}\frac{\pi }{6}\;\;{\rm{and}}\;\;y = 2n\pi \pm \frac{{2\pi }}{3}

Which of the following given values are correct

Option: 1

(i) and (ii)


Option: 2

(ii) and (iii)


Option: 3

(iii) and (iv)


Option: 4

None of these


Answers (1)

best_answer

Here we have a system of 2 equations in 2 variables: x and y

We have,  2^{sinx + cosy} = 1 = 20

\Rightarrow          sinx + cosy = 0                         ..…(i)

Also, given that

16^{ {\sin }^2 x + {\cos }^2y }= 4 = {16^{1/2}}

 \Rightarrow    \sin ^2x + \cos ^2y = \frac{1}{2}            ..…(ii)

Eliminating cos(y) from (i) and (ii), we get

2{\sin ^2}x = \frac{1}{2}   \Rightarrow   \sin x = \pm \frac{1}{2}

Now, when \sin x = \frac{1}{2}     \Rightarrow \cos y = - \frac{1}{2}

 \Rightarrow         x = n\pi + (-1)^n \frac{\pi }{6}   and   y = 2n\pi \pm \frac{{2\pi }}{3}

And  

when   \sin x = - \frac{1}{2}    \Rightarrow   \cos y = \frac{1}{2}

\Rightarrow       x = n\pi + {( - 1)^{n + 1}}\frac{\pi }{6}   and   y = 2n\pi \pm \frac{\pi }{3}

Posted by

manish painkra

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