Or perhaps sin 3x = 3 sin x - 4 \(\sin^{3} x \). Equate 3 sin x - 4 \(\sin^{3} x \) to 0 and we have sin x= 0 or \(\frac{\sqrt{3}} {2} \). Similarly, for sin 3x =0, the basic value for x to satisfy would be \(\frac{\pi}{3} \). This means sin \(\frac{\pi}{3} \) = \(\frac{\sqrt{3}} {2} \) and \(\sin^{-1} \) (\(\frac{\sqrt{3}}{2} \)) =\(\frac{\pi}{3} \) . Now, cos \(\frac{\pi}{3} \) = \(\frac{1}{2} \) since \(\sin^2 x\) + \(\cos^2 x\) =1. It follows that \(cos^{-1} (\frac{1}{2} \))= \(\frac{\pi}{3} \)

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:

TopNewestMaybe first start with cos 3x = 4 \(\cos^{3} \) x - 3 cos x?

Log in to reply

For 4 \(\cos^{3} \) x - 3 cos x=0, we similarly have cos x =0 or \(\frac{\sqrt{3}}{2}\).

cos 3x = 0 means the smallest positive value for 3x is \(\frac{\pi}{2} \) since cos \(\frac{\pi}{2} \) = 0 which follows that x= \(\frac{\pi}{6} \).

We can also prove here that cos \(\frac{\pi}{6} \) = \(\frac{\sqrt{3}}{2}\). Then \(\cos^{-1} (\frac{\sqrt{3}}{2}\) ) = \(\frac{\pi}{6} \).

Log in to reply

I think instead of using the word "smallest positive value" , you should use : the smallest solution in the Principal Interval .

Did you know that \(cos^{-1} cos x = x \forall x \in [0,\pi] \) ?

I have provided you with the graph of \(y=cos^{-1} cos x\)

Log in to reply

And just like what we did for \(\frac{\pi}{3} \), we can also show that \(\sin^{-1} (\frac{1}{2} \)) = \(\frac{\pi}{6} \).

Log in to reply

Or perhaps sin 3x = 3 sin x - 4 \(\sin^{3} x \). Equate 3 sin x - 4 \(\sin^{3} x \) to 0 and we have sin x= 0 or \(\frac{\sqrt{3}} {2} \). Similarly, for sin 3x =0, the basic value for x to satisfy would be \(\frac{\pi}{3} \). This means sin \(\frac{\pi}{3} \) = \(\frac{\sqrt{3}} {2} \) and \(\sin^{-1} \) (\(\frac{\sqrt{3}}{2} \)) =\(\frac{\pi}{3} \) . Now, cos \(\frac{\pi}{3} \) = \(\frac{1}{2} \) since \(\sin^2 x\) + \(\cos^2 x\) =1. It follows that \(cos^{-1} (\frac{1}{2} \))= \(\frac{\pi}{3} \)

Log in to reply

Someone think of a shorter way?

Log in to reply

Log in to reply