Question

How do you solve the expression $2\cos \left( \dfrac{x}{3} \right)+\sqrt{2}=0$?

Answer 1

Hint:We have been given a linear equation in the trigonometric function, cosine of x which also consists of a constant term on the left hand side. Thus, we shall first transpose the constant term to the left hand side. Then we shall divide both sides to make the coefficient of cos x equal to 1. In order the value of x, we will compare with the respective value of the cosine function value.

Complete step by step solution:
Given that $2\cos \left( \dfrac{x}{3} \right)+\sqrt{2}=0$.
We shall first transpose the constant term to the right hand side with a negative sign.
$\Rightarrow 2\cos \left( \dfrac{x}{3} \right)=-\sqrt{2}$
On dividing both sides by 2 to make the coefficient of cos x equal to 1, we get
$\Rightarrow \cos \left( \dfrac{x}{3} \right)=-\dfrac{\sqrt{2}}{2}$
$\Rightarrow \cos \left( \dfrac{x}{3} \right)=-\dfrac{1}{\sqrt{2}}$
We know that $\cos \dfrac{3\pi }{4}=-\dfrac{1}{\sqrt{2}}$ and $\cos \dfrac{5\pi }{4}=-\dfrac{1}{\sqrt{2}}$for its principle solution set, that is, in the interval $\left[ 0,2\pi \right]$.
Comparing this value of cosine function with the calculated value from the given equation, we get
For $\cos \dfrac{3\pi }{4}=-\dfrac{1}{\sqrt{2}}$;
$\Rightarrow \cos \left( \dfrac{x}{3} \right)=\cos \dfrac{3\pi }{4}$
Therefore, we get $\dfrac{x}{3}=\dfrac{3\pi }{4}$.
Here, our aim is to make the coefficient of x equal to 1 in order to obtain a well-defined solution for it. Thus, we shall multiply both sides of the equation with 3.
$\Rightarrow x=\left( 3 \right)\dfrac{3\pi }{4}$
$\Rightarrow x=\dfrac{9\pi }{4}$
For $\cos \dfrac{5\pi }{4}=-\dfrac{1}{\sqrt{2}}$;
$\cos \left( \dfrac{x}{3} \right)=\cos \dfrac{5\pi }{4}$
Therefore, we get $\dfrac{x}{3}=\dfrac{5\pi }{4}$.
Similarly, to make coefficient of x equal to 1, multiplying both sides by 3, we get
$\Rightarrow x=\left( 3 \right)\dfrac{5\pi }{4}$
$\Rightarrow x=\dfrac{15\pi }{4}$
The principle solution for $2\cos \left( \dfrac{x}{3} \right)+\sqrt{2}=0$ is $x=\dfrac{9\pi }{4},\dfrac{15\pi }{4}$.
However, the general solution would be calculated as $\dfrac{9\pi }{4}+2n\pi $ and $\dfrac{15\pi }{4}+2n\pi $ where n is any integer.
Therefore, the solution for the given equation $2\cos \left( \dfrac{x}{3} \right)+\sqrt{2}=0$ is $\dfrac{9\pi }{4}+2n\pi $ and $\dfrac{15\pi }{4}+2n\pi $ where $n\in \mathbb{Z}$.

Note: One possible mistake we could have made was during transposing the constant term. We must observe that if the sign of the constant term would not have been reversed then it would have given a completely different solution set of the given equation which would be $\dfrac{3\pi }{4}+2n\pi $ where $n\in \mathbb{Z}$. Thus, the sign must be reversed carefully during transposition of terms.