trig help please!!!

**ladythugette** · January 19th 2010, 02:06 PM

Solve the equation:

π-3arccosθ=0

all i understand is arccosθ=0 means cos0=θ

but i'm getting confused when there is a value multiplying arc

the answer is arccosθ=π/3,θ=cosπ/3 =1/2

**Archie Meade** · January 19th 2010, 02:13 PM

$\pi-3Cos^{-1}\theta=0$

$\pi=3Cos^{-1}\theta$

$\frac{\pi}{3}=Cos^{-1}\theta$

$Cos^{-1}\ is\ the\ inverse\ operation\ to\ Cos$

$\theta=Cos\frac{\pi}{3}$

**skeeter** · January 19th 2010, 02:13 PM

Originally Posted by ladythugette

Solve the equation:

π-3arccosθ=0

all i understand is arccosθ=0 means cos0=θ

but i'm getting confused when there is a value multiplying arc

the answer is arccosθ=π/3,θ=cosπ/3 =1/2

this is a poorly posed equation ... the input for an inverse trig function is normally a trig ratio, not an angle, because the output of the arccos function is an angle. $\theta$ represents an angle.

using $x$ instead of $\theta$ ...

$\pi - 3\arccos{x} = 0$

$\arccos{x} = \frac{\pi}{3}$

$x = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$

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