The solution set of the equation $\tan x+\sec x=2 \cos x$, in the interval $[0,2 \pi]$ is

The solution set of the equation $\tan x+\sec x=2 \cos x$, in the interval $[0,2 \pi]$ is
  1. $\left\{\frac{\pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}\right\}$
  2. $\left\{\frac{5 \pi}{6}, \frac{7 \pi}{6}, \frac{3 \pi}{2}\right\}$
  3. $\left\{\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}\right\}$
  4. $\left\{\frac{5 \pi}{6}, \frac{11 \pi}{6}, \frac{3 \pi}{2}\right\}$

Solution

Equation: $\tan x+\sec x=2 \cos x$, in the interval $[0,2 \pi]$. Step 1: Substitute $\sec x=\frac{1}{\cos x}$ and $\tan x=\frac{\sin x}{\cos x}$ : $\begin{gathered} \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2 \cos x \\ \frac{\sin x+1}{\cos x}=2 \cos x \end{gathered}$
Step 2: Multiply through by $\cos x$ (assuming $\cos x \neq 0$ ): $\sin x+1=2 \cos ^2 x$
Step 3: Use $\cos ^2 x=1-\sin ^2 x$ : $\begin{gathered} \sin x+1=2\left(1-\sin ^2 x\right) \\ \sin x+1=2-2 \sin ^2 x \end{gathered}$
Step 4: Rearrange into a quadratic in $\sin x$ : $2 \sin ^2 x+\sin x-1=0$
Step 5: Solve the quadratic equation: $\begin{gathered} \sin x=\frac{-1 \pm \sqrt{1-4(2)(-1)}}{2(2)} \\ \sin x=\frac{-1 \pm \sqrt{9}}{4} \\ \sin x=\frac{-1+3}{4} \text { or } \quad \sin x=\frac{-1-3}{4} \\ \sin x=\frac{1}{2} \quad \text { or } \quad \sin x=-1 \end{gathered}$
Step 6: Find solutions in $[0,2 \pi]$ : - For $\sin x=\frac{1}{2}, x=\frac{\pi}{6}, \frac{5 \pi}{6}$. - For $\sin x=-1, x=\frac{3 \pi}{2}$.
Answer: $\left\{\frac{\pi}{6}, \frac{5 \pi}{6}, \frac{3 \pi}{2}\right\}$, Option 3.

Asked in: MHT CET 2024 (15 May Shift 2)

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