Question

The principal solutions of the equation $\sec x+\tan x=2\cos x$ are

• A. $\frac{\pi}{6}$ , $\frac{5\pi}{6}$
• B. $\frac{\pi}{6}$ , $\frac{7\pi}{6}$
• C. $\frac{\pi}{6}$ , $\frac{2\pi}{3}$
• D. $\frac{\pi}{6}$ , $\frac{11\pi}{6}$

Hint:

Trigonometry Tip: Always check the domain of the original equation! Extraneous solutions often arise when squaring sides or cross-multiplying denominators that could equal zero (like $\cos x$ in this problem).

Answer 1

The Correct Option is A

Concept: Trigonometry - Solving Trigonometric Equations for Principal Solutions.

Step 1: Determine the domain restrictions. The given equation contains $\sec x$ and $\tan x$, which are undefined when $\cos x = 0$. Therefore, the solutions cannot be odd multiples of $\frac{\pi}{2}$ (e.g., $x \neq \frac{\pi}{2}, \frac{3\pi}{2}$).

Step 2: Convert the equation entirely into sine and cosine. Rewrite the original equation $\sec x + \tan x = 2\cos x$ in terms of sine and cosine: $\frac{1}{\cos x} + \frac{\sin x}{\cos x} = 2\cos x$. Combining the fractions yields $\frac{1 + \sin x}{\cos x} = 2\cos x$.

Step 3: Cross-multiply and form a quadratic-style equation in sine. Multiply both sides by $\cos x$: $1 + \sin x = 2\cos^2 x$. To solve this, we must express everything in terms of a single trigonometric function. Use the Pythagorean identity $\cos^2 x = 1 - \sin^2 x$ to substitute for $\cos^2 x$: $1 + \sin x = 2(1 - \sin^2 x) = 2 - 2\sin^2 x$.

Step 4: Rearrange into a standard quadratic equation and factor. Move all terms to one side to form a quadratic equation in terms of $\sin x$: $2\sin^2 x + \sin x + 1 - 2 = 0 \implies 2\sin^2 x + \sin x - 1 = 0$. Factor the quadratic equation: $2\sin^2 x + 2\sin x - \sin x - 1 = 0 \implies 2\sin x(\sin x + 1) - 1(\sin x + 1) = 0$. This gives $(2\sin x - 1)(\sin x + 1) = 0$.

Step 5: Find the principal solutions, respecting domain restrictions. From the factored form, we have two possibilities: $\sin x = \frac{1}{2}$ or $\sin x = -1$.
However, if $\sin x = -1$, then $\cos x = 0$ (from $\sin^2 x + \cos^2 x = 1$). From Step 1, we know $\cos x \neq 0$ because it makes the original equation undefined. Thus, we reject $\sin x = -1$.
We are left with $\sin x = \frac{1}{2}$. Since sine is positive in the first and second quadrants, the principal solutions (solutions between $0$ and $2\pi$) are $x = \frac{\pi}{6}$ and $x = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. $$ \therefore \text{The principal solutions are } \frac{\pi}{6} \text{ and } \frac{5\pi}{6}. $$