Solving equation for x

**PleaseHelp** · July 2nd 2011, 04:10 PM

1. arcsin(3x -pi) = 1/2

2. arcsin rad(2x) = arccos rad(x)

**TKHunny** · July 2nd 2011, 04:15 PM

Two things:

1) The argument for arcsin(x) and arccos(x) requires $-1 \le x \le 1$

2) What does "rad(x)" mean?

**PleaseHelp** · July 2nd 2011, 04:25 PM

Originally Posted by TKHunny

Two things:

1) The argument for arcsin(x) and arccos(x) requires $-1 \le x \le 1$

2) What does "rad(x)" mean?

1) Correct...I still don't understand it though. How can an arcsin form an angle of 1/2?

2) I posted the equations again below (in attachment). Little better format.

**skeeter** · July 2nd 2011, 05:17 PM

Originally Posted by PleaseHelp

1) Correct...I still don't understand it though. How can an arcsin form an angle of 1/2?

1/2 is an angle in radians

1. as mentioned earlier, note that ...

$-1 \le 3x-\pi \le 1$

therefore the value of x is restricted to ...

$\frac{\pi - 1}{3} \le x \le \frac{\pi + 1}{3}$

solving the equation ...

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

$x = \frac{1}{3}\left[\sin\left(\frac{1}{2}\right) + \pi\right] \approx 1.207$

2. since arcsin(something) = arccos(something else) , the angle these two expressions equal can only be in quad I (why is that?)

let $\theta = \arcsin{\sqrt{2x}}$

$\cos{\theta} = \sqrt{1-2x}$

since the angles are equal ...

$\theta = \arccos{\sqrt{x}}$

$\cos{\theta} = \sqrt{x}$

so ...

$\sqrt{1-2x} = \sqrt{x}$

solve for $x$

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