Question

The general solution of \(cosθ-sinθ = 1\) is

• A. \(θ = 2nπ-\frac{π}{2}\) or \(θ = 2nπ,n\in Z\)
• B. \(θ = nπ-\frac{π}{2}\) or \(θ = nπ,n\in Z\)
• C. \(θ = n\frac{π}{2}-π\) or \(θ = n\frac{π}{2},n\in Z\)
• D. \(θ = 2nπ+\frac{π}{2}\) or \(θ = nπ,n\in Z\)

Hint:

Write \(\cos\theta-\sin\theta = \sqrt2\cos(\theta+\pi/4)\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We rewrite the left side as a single cosine using the identity \(a\cos\theta + b\sin\theta = R\cos(\theta-\phi)\), with \(R = \sqrt{a^2+b^2}\).

Step 2: Convert:
\[ \cos\theta-\sin\theta = \sqrt2\left(\frac{1}{\sqrt2}\cos\theta - \frac{1}{\sqrt2}\sin\theta\right) = \sqrt2\cos\left(\theta+\frac{\pi}{4}\right) \]

Step 3: Solve:
\(\sqrt2\cos(\theta+\frac{\pi}{4}) = 1\) gives \(\cos(\theta+\frac{\pi}{4}) = \frac{1}{\sqrt2} = \cos\frac{\pi}{4}\).
So \(\theta + \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4}\).
With the plus sign: \(\theta = 2n\pi\). With the minus sign: \(\theta = 2n\pi - \frac{\pi}{2}\).

Step 4: Other options:
Options B, C and D use \(n\pi\) or \(\frac{n\pi}{2}\) and so include angles such as \(\pi\), where \(\cos\pi - \sin\pi = -1 \ne 1\). They are wrong.

Final Answer:
The general solution is \(\theta = 2n\pi\) or \(2n\pi - \frac{\pi}{2}\), option (A). \[ \boxed{\theta = 2n\pi-\frac{\pi}{2}\ \text{or}\ \theta = 2n\pi} \]