Question

How do you solve $\cos x+\sin x\tan x=2$? \[\]

Answer 1

Hint: We convert the tangent function in the given trigonometric equation into sine and cosine using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and then we use the Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. We find trigonometric equation only cosine of the form $\cos x=\cos \alpha $ where $\alpha $ is principal solution whose general solution is given by $x=2n\pi \pm \alpha $.\[\]

Complete step by step answer:
We know that a trigonometric equation is an equation with trigonometric functions with unknown arguments as measure of angles. When we are asked to solve a trigonometric equation we have to find all possible measures of unknown angles.
We know that the first solution of the trigonometric equation within the interval $\left[ 0,2\pi \right]$ is called principal solution and using periodicity all possible solutions obtained with integer $n$ are called general solutions. The general solution of the trigonometric equation $\cos \theta = \cos \alpha $ with principal solution $\theta =\alpha $ are given by
\[\theta = 2n\pi \pm \alpha \]
Here $n$ is any integer may be negative, positive or zero. We are given the following trigonometric equation in the question
\[\cos x+\sin x\tan x=2\]
 We convert the tangent function in the above equation into sine and cosine using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ in the above step to have;
\[\begin{align}
  & \Rightarrow \cos x+\sin x\dfrac{\sin x}{\cos x}=2 \\
 & \Rightarrow \cos x+\dfrac{{{\sin x }^{2}}}{\cos x}=2 \\
 & \Rightarrow \dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\cos x}=2 \\
\end{align}\]
We use the Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the above step to have ;
\[\begin{align}
  & \Rightarrow \dfrac{1}{\cos x}=2 \\
 & \Rightarrow 1=2\cos x \\
 & \Rightarrow \cos x=\dfrac{1}{2} \\
\end{align}\]
We know that from basic trigonometric value table that the first angel for which cosine value is $\dfrac{1}{2}$ is ${{60}^{\circ }}=\dfrac{\pi }{3}$. So we have
\[\Rightarrow \cos x=\cos \left( \dfrac{\pi }{3} \right)\]
So we have the principal solution $\alpha =\dfrac{\pi }{3}$. So the general solution of the trigonometric is given by
\[x=2n\pi \pm \alpha =2n\pi \pm \dfrac{\pi }{3}\]

Note: We note that $\tan x$ is not defined for $x=\left( 2m+1 \right)\dfrac{\pi }{2}$ for any integer$m$. Let us see if there are solutions not valid because of the definition of $\tan x$. We have
\[\begin{align}
  & 2n\pi \pm \dfrac{\pi }{3}=\left( 2m+1 \right)\dfrac{\pi }{2} \\
 & \Rightarrow 2n\pm \dfrac{1}{3}=\dfrac{2m+1}{2} \\
 & \Rightarrow \dfrac{6n\pm 1}{3}=\dfrac{2m+1}{2} \\
\end{align}\]
We cross multiply to have ;
\[\begin{align}
  & \Rightarrow 12n\pm 2=6m+3 \\
 & \Rightarrow 12n-6m=3\mp 2 \\
 & \Rightarrow 6\left( 2n-m \right)=5,1 \\
 & \Rightarrow 2n-m=\dfrac{5}{6},2n-m=\dfrac{1}{6} \\
\end{align}\]
Since multiplication and addition are closed within the set of integers $2n-m$ have to be integers. So there does not exist $n,m\in Z$ such that the solution $2n\pi \pm \dfrac{\pi }{3}=\left( 2m+1 \right)\dfrac{\pi }{2}$. Hence our solution is valid.