The general solution of the equation {3} θ+ θ={2} is

Correct Answer is : nπ+(−1)nπ4−π3,n∈Z{n} π+(-1)^{{n}} π/4-π/3, {n} {Z}nπ+(−1)n4π−3π,n∈Z

{n} π+(-1)^{{n}} π/2+π/6, {n} {Z}

{n} π+(-1)^{{n}} π/2-π/6, {n} {Z}

{n} π+(-1)^{{n}} π/4-π/3, {n} {Z}

{n} π+(-1)^{{n}} π/4+π/3, {n} {Z}

Correct Answer is : nπ+(−1)nπ4−π3,n∈Z{n} π+(-1)^{{n}} π/4-π/3, {n} {Z}nπ+(−1)n4π−3π,n∈Z

Test Index

MHTCET 16 May 2024 Shift 1 Paper

Question : 109 of 150

Marks: +1, -0

\sqrt{3} \cos \theta+\sin \theta=\sqrt{2}

$\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{2}+\frac{\pi}{6}, \mathrm{n} \in \mathbb{Z}$
$\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{2}-\frac{\pi}{6}, \mathrm{n} \in \mathbb{Z}$
$\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}-\frac{\pi}{3}, \mathrm{n} \in \mathbb{Z}$
$\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{4}+\frac{\pi}{3}, \mathrm{n} \in \mathbb{Z}$

Solution:

Concept:

Use the compound-angle identity

\sin(A+B)=\sin A\cos B+\cos A\sin B

to combine the terms into a single sine function.

Explanation:

Given equation:

\sqrt{3}\cos\theta+\sin\theta=\sqrt{2}

Divide both sides by

2

\frac{\sqrt{3}}{2}\cos\theta+\frac{1}{2}\sin\theta=\frac{\sqrt{2}}{2}

Recognize

\frac{\sqrt{3}}{2}=\sin\frac{\pi}{3}

and

\frac{1}{2}=\cos\frac{\pi}{3}

So the left side becomes:

\sin\frac{\pi}{3}\cos\theta+\cos\frac{\pi}{3}\sin\theta=\sin\left(\theta+\frac{\pi}{3}\right)

And

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

Therefore,

\sin\left(\theta+\frac{\pi}{3}\right)=\sin\frac{\pi}{4}

The general solution for

\sin x=\sin\alpha

is:

x=n\pi+(-1)^n\alpha

n\in\mathbb{Z}

Thus,

\theta+\frac{\pi}{3}=n\pi+(-1)^n\frac{\pi}{4}

So,

\theta=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{3}

n\in\mathbb{Z}

Answer:

\theta=n\pi+(-1)^n\frac{\pi}{4}-\frac{\pi}{3}

n\in\mathbb{Z}

This matches option C.

Go to Question:

Prev Question

Next Question