Question

Solve: $\left| \cos x \right|=\cos x-2\sin x$
[a] $x=\left( 2n+1 \right)\pi +\dfrac{\pi }{4}$
[b] $\left( 2n+1 \right)\pi +\dfrac{\pi }{2}$
[c] $2n\pi +\dfrac{\pi }{4}$
[d] $2n\pi +\dfrac{\pi }{2}$

Answer 1

Hint: Use the fact that if $\left| x \right|=a,a\ge 0$, then the solution of the given system is $x=\pm a$. Hence the system $\left| \cos x \right|=\cos x-2\sin x$, will have a solution if $\cos x-2\sin x\ge 0$ and the solution of the system will be $\cos x=\pm \left( \cos x-2\sin x \right)$. Take the signs individually and solve the equations. Use the fact that if tanx = tany, then $x=n\pi +y,n\in \mathbb{Z}$ and if sinx = siny, then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$. Remove the extraneous roots and hence find the general solution of the equation. Alternatively, find the roots of the given equation in the interval $\left[ 0,2\pi \right)$. If the set of the roots found is S(say), then the general solution of the given equation is x\in \left\\{ 2n\pi +y,n\in \mathbb{Z},y\in S \right\\}.

Complete step-by-step answer:
We have $\left| \cos x \right|=\cos x-2\sin x$.
Since $\left| x \right|\ge 0,$ we have $\cos x-2\sin x\ge 0\text{ }\left( a \right)$
The equation (a) will be later on used.
Now, we know that if $\left| x \right|=a,a\ge 0$, then the solution of the given system is $x=\pm a$.
Hence, we have
$\cos x=\pm \left( \cos x-2\sin x \right)$
Taking the positive sign, we get
$\cos x=\cos x-2\sin x\Rightarrow 2\sin x=0\Rightarrow \sin x=0$
Taking the negative sign, we get
$\cos x=-\cos x+2\sin x\Rightarrow 2\cos x=2\sin x\Rightarrow \tan x=1$
Solving sinx = 0:
We know that sin(0) = 0
Hence, we have sinx = sin0
We know that if sinx = siny, then $x=n\pi +{{\left( -1 \right)}^{n}}y,n\in \mathbb{Z}$
Hence, we have
$x=n\pi ,n\in \mathbb{Z}$
Solving tanx = 1
We have
$\tan \left( \dfrac{\pi }{4} \right)=1$
Hence, we have
$\tan x=\tan \dfrac{\pi }{4}$
We know that of tanx = tany, then $x=n\pi +y,n\in \mathbb{Z}$
Hence, we have
$x=n\pi +\dfrac{\pi }{4},n\in \mathbb{Z}$
Now consider the solution $x=n\pi ,n\in \mathbb{Z}$
When n is odd, we have $\cos x=\cos \left( \left( 2k+1 \right)\pi \right)=-1$ and $\sin \left( \left( 2k+1 \right)\pi \right)=1$
Hence, we have
$\cos x < 2\sin x$
Hence when n is odd $x=n\pi$ does not satisfy equation(a) and hence is not the solution of the given equation.
When n is even, we have $\cos x=\cos 2k\pi =1$ and $\sin x=\sin 2k\pi =0$
Hence, we have $\cos x\ge 2\sin x$ and hence when n is even $x=n\pi$ is a solution of the given equation
Consider the solution $x=n\pi +\dfrac{\pi }{4}$
When n is odd, we have $\cos x=-\dfrac{1}{\sqrt{2}}$ and $\sin x=\dfrac{-1}{\sqrt{2}}$
Hence, we have $\cos x\ge 2\sin x$
Hence when n is odd, we have $x=n\pi +\dfrac{\pi }{4}$ is a solution of the given equation
When n is even, we have $\cos x=\dfrac{1}{\sqrt{2}}$ and $\sin x=\dfrac{1}{\sqrt{2}}$
Hence, we have $\cos x<2\sin x$
Hence, when n is even, $x=n\pi +\dfrac{\pi }{4}$ is not a solution of the given equation.
Hence the solution set of the given equation is \left\\{ 2k\pi ,k\in \mathbb{Z} \right\\}\bigcup \left\\{ \left( 2k+1 \right)\pi +\dfrac{\pi }{4},k\in \mathbb{Z} \right\\}=\left\\{ 2k\pi ,k\in \mathbb{Z} \right\\}\bigcup \left\\{ 2k\pi +\dfrac{3\pi }{4},k\in \mathbb{Z} \right\\}, which is the required solution set of the given equation.
Hence option [a] is correct.

Note: Alternative method:
In the interval $\left[ 0,2\pi \right)$, we have $x=0,x=\dfrac{3\pi }{4}$ satisfy the given equation.
Hence, we have the solution of the given equation is x\in \left\\{ 2k\pi ,2k\pi +\dfrac{\pi }{4},k\in \mathbb{Z} \right\\}