MemeLord

Nevermind I found it using the search feature. It was just AdamTaurus. Now time to cycle through the 40 passwords I've used over the years to see which one I used 

Ok, thank you very much CPhill.

When I did it on Desmos (https://www.desmos.com/calculator/riwpauxdrs), I got a few solutions but I don't know what they are in radian format.

1) \(sin(2x)-sin(x)cos(x)=cos(x)\)

For the sin (2x), you can use the Double Angle Formula of \(sin(2x)=2sin(x)cos(x)\).

\(2sin(x)cos(x)-sin(x)cos(x)=cos(x)\)

Then the \(2sin(x)cos(x)-sin(x)cos(x)\) simplifies to \(sin(x)cos(x)\).

\(sin(x)cos(x)=cos(x)\)

Then subtract cos(x) from both sides.

\(sin(x)cos(x)-cos(x)=0\)

You can factor the cos(x) out.

\(cos(x)(sin(x)-1)=0\)

Then, set both terms equal to 0.

\(cos(x)=0,sin(x)-1=0\)

Add 1 to both sides on the sin(x) function.

\(cos(x)=0,sin(x)=1\)

Then, take the inverse of both functions.

\(x=cos^{-1}(0),x=sin^{-1}(1)\)

The only values that satisfy those are \(x=90^o, x=270^o\) or \(x=\frac{\pi}{2},x=\frac{3\pi}{2}\).

10.47 should be sufficient since you aren't finding all possible values of the sine function and there isn't a squared anywhere.

\(\frac{sin(33.2)}{x}=\frac{sin(45.6)}{13.7}\)

Alright, so the first thing you want to do here is to find the values of the sine functions.

\(sin(33.2) = 0.547563223493\)

Which rounded to 2 decimal places is 0.55.

\(sin(45.6) = 0.714472679633​\)

Rounded to 2 decimal places that's 0.72 (the 7 behind the two 4's carries down the line).

So, now we have: \(\frac{0.55}{x}=\frac{0.72}{13.7}\)

Then you have to cross multiply.

\(0.72x=(13.7)(0.55)\)

Then multiply 13.7 by 0.55.

\((13.7)(0.55) = 7.535​\)

Round to 7.54

\(0.72x=7.54\)

Then divide both sides by 0.72.

\(7.54/0.72 = 10.4722222222222222\)

Round that to 10.47

\(x=10.47\)

Does that make sense?

First of all, to find the inverse, you have to switch the x and y values and solve for y.

So \(y=sin^{-1}(3x)\) becomes \(x=sin^{-1}(3y)\).

To begin solving for y, you must take the sine of both sides.

\(sin(x)=sin(sin^{-1}(3y))\)

The sin of sin^-1 cancel.

\(sin(x)=3y\)

Then you divide both sides by 3.

\(\frac{sin(x)}{3}=\frac{3y}{3}\)

The 3's on the y side cancel.

\(y=\frac{sin(x)}{3}\)

So, the inverse of \(f(x)=sin^{-1}(3x)\) is \(f^{-1}(x)=\frac{sin(x)}{3}\).
 

By graphing, you can find the domain and range of f^-1(x).

So, the domain of \(f^{-1}=\frac{sin(x)}{3}\) is \(-\infty < x < \infty\)  , and the range is  \(-\frac{\pi}{2}\leq y\leq\frac{\pi}{2}\).

Is the unique solution supposed to be \(re^{i\theta}\) or \(re^{\frac{i}{\theta}}\)?

Close enough I guess.

Is there a way to do it using like pythagorean identities and stuff like that? That's the way my professor wants us to do it.