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Find the number of solutions of the equation given below,

\(\sqrt{3}\sin \theta+ \cos \theta = 4\)

Note by Akshay Yadav 2 years ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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For real solutions? Answer is 0.

Just apply the Trigonometric R method.

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Thank you very much!

No, this is the quickest method. \(\quad\quad \quad\)

Since the maximum real value for \(\sin \theta\) and \(\cos \theta\) is \(1\), LHS cannot be larger than \(\sqrt{3}+1 < 4\), therefore, there is no real root.

Your method is right, but the maximum value of LHS is not \(\sqrt3+1\).

I did not say that the maximum value is \(\sqrt{3}+1\). What I meant was LHS \(\ne\) RHS.

Thanks! I also used the same method to solve the question in exam and wondering if it was correct or not.

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestFor real solutions? Answer is 0.

Just apply the Trigonometric R method.

Log in to reply

Thank you very much!

Log in to reply

No, this is the quickest method. \(\quad\quad \quad\)

Log in to reply

Since the maximum real value for \(\sin \theta\) and \(\cos \theta\) is \(1\), LHS cannot be larger than \(\sqrt{3}+1 < 4\), therefore, there is no real root.

Log in to reply

Your method is right, but the maximum value of LHS is not \(\sqrt3+1\).

Log in to reply

I did not say that the maximum value is \(\sqrt{3}+1\). What I meant was LHS \(\ne\) RHS.

Log in to reply

Thanks! I also used the same method to solve the question in exam and wondering if it was correct or not.

Log in to reply