Question

The most general solution of the system $\tan x=-1$ and $\cos x=\dfrac{1}{\sqrt{2}}$ is
[a] $n\pi +\dfrac{7\pi }{4},n\in \mathbb{Z}$
[b] $n\pi +{{\left( -1 \right)}^{n}}\dfrac{7\pi }{4},n\in \mathbb{Z}$
[c] $2n\pi +\dfrac{7\pi }{4},n\in \mathbb{Z}$
[d] None of these

Answer 1

Hint: Use the fact that the general solution of the equation tanx = tany is $x=n\pi +y,n\in \mathbb{Z}$ and the general solution of the equation cosx=cosy is $x=2n\pi \pm y,n\in \mathbb{Z}$. Use the fact that $\tan \left( \dfrac{3\pi }{4} \right)=-1$ and $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$.

Complete step-by-step answer:
Hence find the general solution of the system.
We have
tanx = -1
Hence, we have
$\tan x=\tan \left( \dfrac{3\pi }{4} \right)$
We know that the general solution of the equation tanx = tany is given by $x=n\pi +y,n\in \mathbb{Z}$
Hence, we have
$x=n\pi +\dfrac{3\pi }{4},n\in \mathbb{Z}$
Also, we have
$\cos x=\dfrac{1}{\sqrt{2}}$
Hence, we have
$\cos x=\cos \dfrac{\pi }{4}$
We know that the general solution of the equation cosx = cosy is given by $x=2n\pi \pm y,n\in \mathbb{Z}$
Hence, we have
$x=2n\pi \pm \dfrac{\pi }{4}$
Now, not that $\left\{ n\pi +\dfrac{3\pi }{4},n\in \mathbb{Z} \right\}=\left\{ n\pi +\pi +\dfrac{3\pi }{4},n\in \mathbb{Z} \right\}=\left\{ n\pi +\dfrac{7\pi }{4},n\in \mathbb{Z} \right\}$( since replacing n by n -1 will not change the solutions in the given set)
Similarly, we have
$\left\{ 2n\pi \pm \dfrac{\pi }{4},n\in \mathbb{Z} \right\}=\left\{ 2n\pi \pm \left( 2\pi -\dfrac{\pi }{4} \right),n\in \mathbb{Z} \right\}=\left\{ 2n\pi \pm \dfrac{7\pi }{4},n\in \mathbb{Z} \right\}$
The above two transformations can be understood as follows:
Instead of using $\tan \left( \dfrac{3\pi }{4} \right)=-1$, we are using $\tan \left( \dfrac{7\pi }{4} \right)=-1$ , and instead of using $\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$, we are using $\cos \dfrac{7\pi }{4}=\dfrac{1}{\sqrt{2}}$. The solution set should remain unchanged, and hence the above two results are obtained.
Hence, we have
$x\in \left\{ n\pi +\dfrac{7\pi }{4},n\in \mathbb{Z} \right\}\bigcap \left\{ 2n\pi \pm \dfrac{7\pi }{4},n\in \mathbb{Z} \right\}=\left\{ 2n\pi +\dfrac{7\pi }{4},n\in \mathbb{Z} \right\}$
This is because any solution of form $n\pi +\dfrac{7\pi }{4}$, which is in $\left\{ 2n\pi \pm \dfrac{7\pi }{4} \right\}$implies that n is an even number.
Hence n = 2m
Hence the solution is of the form of $2m\pi +\dfrac{7\pi }{4}$ and hence the above result.
Hence option [c] is correct.

Note: Alternative solution:
Since tanx is negative, we have x is in the fourth or the second quadrant.
Since cosx is positive, we have x is in the first quadrant or the fourth quadrant.
Hence, we have x is in the fourth quadrant.
The solution in $\left[ 0,2\pi \right]$ of the given system is $x=\dfrac{7\pi }{4}$
Since the common period of cosx and tanx is $2\pi $, we have
The general solution of the system is $x=2n\pi +\dfrac{7\pi }{4},n\in \mathbb{Z}$, which is the same as obtained above.