Question

If \(3\cos x \ne 2\sin x\), then the general solution of \[ \sin^2 x-\cos 2x=2-\sin 2x \] is

• A. \(x=n\pi+\dfrac{\pi}{2},\; n\in\mathbb{Z}\)
• B. \(x=n\left(\dfrac{\pi}{2}\right)+\pi,\; n\in\mathbb{Z}\)
• C. \(x=n\left(\dfrac{\pi}{2}\right)+\dfrac{\pi}{3},\; n\in\mathbb{Z}\)
• D. \(x=(2n+1)\pi,\; n\in\mathbb{Z}\)

Hint:

When a condition is given with a trigonometric equation, always check and reject invalid solutions.

Answer 1

The Correct Option is A

Step 1: Use trigonometric identities.
We know \[ \sin^2 x=\frac{1-\cos 2x}{2} \] Substitute into the given equation: \[ \frac{1-\cos 2x}{2}-\cos 2x=2-\sin 2x \]
Step 2: Simplify the equation.
\[ \frac{1-3\cos 2x}{2}=2-\sin 2x \] Multiply both sides by 2: \[ 1-3\cos 2x=4-2\sin 2x \]
Step 3: Rearrange terms.
\[ 3\cos 2x-2\sin 2x= -3 \] or \[ 2\sin 2x-3\cos 2x=3 \]
Step 4: Write in standard form.
\[ \sqrt{(2)^2+(3)^2}\sin(2x-\alpha)=3 \] where \[ \sin\alpha=\frac{3}{\sqrt{13}},\quad \cos\alpha=\frac{2}{\sqrt{13}} \] Hence, \[ \sqrt{13}\sin(2x-\alpha)=3 \Rightarrow \sin(2x-\alpha)=\frac{3}{\sqrt{13}} \]
Step 5: Solve for \(x\).
This gives \[ 2x-\alpha=\frac{\pi}{2}+2n\pi \Rightarrow x=n\pi+\frac{\pi}{2} \]
Step 6: Check the given condition.
The condition \(3\cos x\ne2\sin x\) removes extraneous solutions, leaving \[ x=n\pi+\frac{\pi}{2} \]