The general solution of θ- θ=1 is

Correct Answer is : θ=2nπ−π2θ=2 n π-π/2θ=2nπ−2π or θ=2nπ,n∈Zθ=2 n π, n Zθ=2nπ,n∈Z

θ=2 n π-π/2 or θ=2 n π, n Z

Correct Answer is : θ=2nπ−π2θ=2 n π-π/2θ=2nπ−2π or θ=2nπ,n∈Zθ=2 n π, n Zθ=2nπ,n∈Z

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MHTCET 13 Apr 2026 Shift 2 Paper

Question : 105 of 150

Marks: +1, -0

The general solution of

\cos \theta-\sin \theta=1

Solution:

Concept:

Use the identity

\cos \theta - \sin \theta = \sqrt{2} \cos\left(\theta + \frac{\pi}{4}\right)

and then solve the cosine equation.

Explanation:

Given:

\cos \theta - \sin \theta = 1

Rewrite using the standard identity:

\sqrt{2} \cos\left(\theta + \frac{\pi}{4}\right) = 1

Divide both sides by

\sqrt{2}

\cos\left(\theta + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}

We know that

\cos x = \frac{1}{\sqrt{2}}

when

x = 2n\pi \pm \frac{\pi}{4}

n \in \mathbb{Z}

Here,

x = \theta + \frac{\pi}{4}

So,

\theta + \frac{\pi}{4} = 2n\pi + \frac{\pi}{4}

\theta + \frac{\pi}{4} = 2n\pi - \frac{\pi}{4}

Solving these gives:

\theta = 2n\pi

\theta = 2n\pi - \frac{\pi}{2}, \quad n \in \mathbb{Z}

Answer:

\theta = 2n\pi - \frac{\pi}{2} \ \text{or} \ \theta = 2n\pi, \quad n \in \mathbb{Z}

Hence, the correct option is Option A.

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