Linear Trig Functions help please!

**Jane Zhang** · December 5th 2007, 05:22 PM

hey! can anybody please help me solve this question please!
2Cos(x/2)=square root two

**Jane Zhang** · December 5th 2007, 05:24 PM

or this question:
5cos3x=-3
thank-you!please reply!

**Soroban** · December 5th 2007, 05:49 PM

Hello, Jane!

$2\cos\left(\frac{x}{2}\right)\;=\;\sqrt{2}$

Divide by 2: . $\cos\left(\frac{x}{2}\right) \:=\:\frac{\sqrt{2}}{2}$

Now what angle has a cosine of $\frac{\sqrt{2}}{2}$ ?
. . That's right . . . $\pm\frac{\pi}{4} + 2\pi n$

So we have: . $\frac{x}{2} \:=\:\frac{\pi}{4} + 2\pi n$

Therefore: . $\boxed{x \:=\:\frac{\pi}{2} + 4\pi n}$

$5\cos(3x)\:=\:-3$

We have: . $\cos(3x) \:=\:\text{-}\frac{3}{5}$

Then: . $3x \:=\:\cos^{\text{-}1}\left(\text{-}\frac{3}{5}\right)$

Therefore: . $\boxed{x \;=\;\frac{1}{3}\cos^{\text{-}1}\left(\text{-}\frac{3}{5}\right)}$

**Jane Zhang** · December 6th 2007, 02:12 PM

thanks a lot!

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