How do you solve $\cos x+\sin x\tan x=2$ over the interval 0 to $2\pi $ ?

Hint: To solve this question first we will simplify the given expression by using trigonometric identities and formulas. We will find the value of x by simplifying the given expression. Then we will find the values of x in the given range.

Complete step by step answer:
We have been given an expression $\cos x+\sin x\tan x=2$.
We have to solve the given expression over the interval 0 to $2\pi $.
The given expression is $\cos x+\sin x\tan x=2$
Now, we know that $\tan x=\dfrac{\sin x}{\cos x}$
Substituting the value in the above expression we will get
$\Rightarrow \cos x+\sin x\dfrac{\sin x}{\cos x}=2$
Now, simplifying the above obtained equation we will get
$\Rightarrow \dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\cos x}=2$
Now, we know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
Substituting the value in the above equation we will get
$\Rightarrow \dfrac{1}{\cos x}=2$
Now, simplifying the above obtained equation we will get
$\begin{align}
  & \Rightarrow \dfrac{1}{2}=\cos x \\
 & \Rightarrow \cos x=\dfrac{1}{2} \\
\end{align}$
Now, we have given the interval 0 to $2\pi $.
We know that cosine function has positive value in the first and fourth quadrant.
We know that $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$
So we get $x=\dfrac{\pi }{3}$.
Now, take $x=\dfrac{\pi }{3}$ as reference angle to calculate the value of fourth quadrant angle we will get
$\Rightarrow 2\pi -\dfrac{\pi }{3}=\dfrac{5\pi }{3}$
Hence we get the value $x=\dfrac{\pi }{3},\dfrac{5\pi }{3}$ over the interval 0 to $2\pi $.

Note:
Students may consider all values of x for cosine function over the interval 0 to $2\pi $, which is incorrect. Here in this question when we solve the expression we get the positive value as $\dfrac{1}{2}$ so we need to consider only positive values of cosine function.