The general solution of x+ x=1 is

Correct Answer is : x=nπ+(−1)nπ4−π4,n∈Zx={n} π+(-1)^{{n}} π/4-π/4, {n} {Z}x=nπ+(−1)n4π−4π,n∈Z

x={n} π+(-1)^{{n}} π/4-π/4, {n} {Z}

Correct Answer is : x=nπ+(−1)nπ4−π4,n∈Zx={n} π+(-1)^{{n}} π/4-π/4, {n} {Z}x=nπ+(−1)n4π−4π,n∈Z

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MHTCET 11 May 2024 Shift 2 Paper

Question : 125 of 150

Marks: +1, -0

The general solution of

\sin x+\cos x=1

$x=2 \mathrm{n} \pi, \mathrm{n} \in \mathbb{Z}$
$x=\mathrm{n} \pi+\frac{\pi}{2}, \mathrm{n} \in \mathbb{Z}$
$x=\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}-\frac{\pi}{4}, \mathrm{n} \in \mathbb{Z}$
not existing

Solution:

Concept:

Use the identity

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

to convert the equation into a standard cosine equation.

Explanation:

Given:

\sin x+\cos x=1

Divide both sides by

\sqrt{2}

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

Use

\frac{1}{\sqrt{2}}=\sin\frac{\pi}{4}=\cos\frac{\pi}{4}

\cos\left(x-\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}

\therefore \cos\left(x-\frac{\pi}{4}\right)=\cos\frac{\pi}{4}

For

\cos\theta=\cos\alpha

, the general solution is

\theta=2n\pi\pm\alpha

So,

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

This gives:

x=2n\pi \quad \text{or} \quad x=2n\pi+\frac{\pi}{2}

Therefore, the combined general solution is:

x=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{4}, \; n\in\mathbb{Z}

Answer:

Option C.

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