Question

The general solution of \(cosx+sinx = cos2x+sin2x\) is \(x = npπ\) or \(x = \frac{nqπ}{3}+\frac{π}{6}\) for \(n\in Z\) then \(p:q =\)

• A. \(1:1\)
• B. \(1:2\)
• C. \(2:3\)
• D. \(2:1\)

Hint:

Group the equation as cos x - cos 2x = sin 2x - sin x and factor.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Use the sum-to-product formulas to factorise the equation and solve each factor.

Step 2: Rearrange:
\(\cos x - \cos 2x = \sin 2x - \sin x\).
Left side: \(\cos x - \cos 2x = 2\sin\frac{3x}{2}\sin\frac{x}{2}\).
Right side: \(\sin 2x - \sin x = 2\cos\frac{3x}{2}\sin\frac{x}{2}\).

Step 3: Solve:
\[ 2\sin\frac{x}{2}\left(\sin\frac{3x}{2} - \cos\frac{3x}{2}\right) = 0 \]
Case 1: \(\sin\frac{x}{2} = 0\) gives \(\frac{x}{2} = n\pi\), so \(x = 2n\pi\).
Case 2: \(\tan\frac{3x}{2} = 1\) gives \(\frac{3x}{2} = n\pi + \frac{\pi}{4}\), so \(x = \frac{2n\pi}{3} + \frac{\pi}{6}\).

Step 4: Compare with the given form:
\(x = np\pi\) means \(p = 2\). \(x = \frac{nq\pi}{3} + \frac{\pi}{6}\) means \(q = 2\). So \(p : q = 2 : 2 = 1 : 1\).

Step 5: Why the other options are wrong.
Ratios 1:2, 2:3 and 2:1 would need \(q\) or \(p\) to take values other than 2, which does not match the solution set above.

Final Answer:
The ratio is \(1:1\), option (A). \[ \boxed{1:1} \]