Question

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

Answer 1

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;

& \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{aligned}$$ We use the Pythagorean trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ in the above step to have ; $$\begin{aligned} & \Rightarrow \dfrac{1}{\cos x}=2 \\\ & \Rightarrow 1=2\cos x \\\ & \Rightarrow \cos x=\dfrac{1}{2} \\\ \end{aligned}$$ We know that from basic trigonometric value table that the first angel for which cosine value is $\dfrac{1}{2}$ is ${{60}^{\circ }}=\dfrac{\pi }{3}$. So we have $$\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{aligned} & 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{aligned}$$ We cross multiply to have ; $$\begin{aligned} & \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{aligned}$$ 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.