Question

The general solutions of $\sin^2x\cdot\sec x=\tan x-\sin x+1$ are

• A. $x=n\pi+(-1)^n\frac{\pi}{4}$ or $x=m\pi+\frac{3\pi}{4},\ m,n\in\mathbb{Z}$
• B. $x=n\pi+(-1)^n\frac{\pi}{2}$ or $x=m\pi+\frac{3\pi}{4},\ m,n\in\mathbb{Z}$
• C. $x=n\pi+(-1)^n\frac{\pi}{2}$ or $x=m\pi+\frac{5\pi}{4},\ m,n\in\mathbb{Z}$
• D. $x=n\pi+(-1)^n\frac{\pi}{4}$ or $x=m\pi+\frac{5\pi}{4},\ m,n\in\mathbb{Z}$

Hint:

Always check for values that make denominators zero when solving trigonometric equations.

Answer 1

The Correct Option is B

Step 1: Simplifying the given equation.
\[ \sin^2x\cdot\sec x=\frac{\sin^2x}{\cos x} \] The equation becomes \[ \frac{\sin^2x}{\cos x}=\tan x-\sin x+1 \]
Step 2: Writing all terms in sine and cosine.
\[ \frac{\sin^2x}{\cos x}=\frac{\sin x}{\cos x}\sin x \] Rearranging, we obtain solutions when \[ \cos x=0 \quad \text{or} \quad \tan x=\sin x \]
Step 3: Solving the cases.
\[ \cos x=0 \Rightarrow x=n\pi+(-1)^n\frac{\pi}{2} \] \[ \tan x=\sin x \Rightarrow x=m\pi+\frac{3\pi}{4} \]
Step 4: Conclusion.
The general solutions are \[ x=n\pi+(-1)^n\frac{\pi}{2}\ \text{or}\ x=m\pi+\frac{3\pi}{4} \]