How can I solve this by hand?

How can I solve this by hand?

By using your hand to rev up a grapher.

you can reduce it to

[math]cos(x)cos(x- arctan(B/A)) = C/ \sqrt{A^2 + B^2} [/math]

you don't pay me enough to give you more.

Solve for what you fucking retard?

>4 dimensional graphing

for B
A,C,x given

For x, ever taken a math class?

By learning trigonometric identities.

for x obviously you brainlet

i'm actually curious myself because I can't come up with a solution

is the first term supposed to be cos(x^2) or cos^2(x)?

You trying to troll or what?

>wow look im so smart i put the equation into a calculator so impressive

cos^2(x)

he's probably smarter than a moron who tries to solve anything by hand if a computer can do it.

>I taught a monkey how to input equations into a calculator so that monkey is smarter than humans

>I taught a monkey to solve equations by hand
>I also taught the same monkey to input equations in a solver
>I presented a monkey with a problem. He made the optimal choice to input it in a fast calculator instead of solving it by hand.

yeah, the monkey is smarter than you.

Of course I can solve it, but I won't bother doing it. Do you see what a clusterfuck the solution is?

Thank you
I just don't really see how you got there

op wants to find the roots you retard, you did nothing

The same way you would solve it with a computer

and I want a different lady to empty my balls every morning you retard. You did nothing.

this is high school math, if you cant solve this then consider sending your cv to mcdonalds today

You're mad because you don't know how to use ti

Substitute u=cos(x) and sin(x)=sqrt(1-u^2). Turn into a 4th degree polynomial, then substitute s=u^2 to turn into a quadratic.

Convert all trig functions to complex exponentials.
cos(x)=(e^(ix)+e^(-ix))/2
sin(x)=(e^(ix)-e^(-ix))/(2i)
You should end up with a quartic equation in e^(ix).

this has to be bait, right? a guy pretending to be retarded so we can all have a good laugh about it.

Thank you, this should both work fine. I also found a way to solve it with identities for Sin*Cos and Cos^2

you're just illustrating your own stupidity by revealing that you think that the poster intended his picture to look complicated, so you necessarily think that what he posted is complicated