sin(x)^2+cos(x)^2=C How do i solve for x?

>>8817137
x can literally be anything because C is always 1.

>how do i solve the idiot formula

Define a function $f(x)=\sin^2(x)+\cos^2(x)$ and take the derivative. Turns out that the derivative is zero, which means that the function is constant. Taking an arbitary $x$ solves the "equation" for $C$. Another approach would be to not be a brainlet and apply the popular formula...

>>8817139
actually, not very often

>>8817137
if C = 1, then solution is all R (or C if you solve in the complex numbers)
if C =/= 1, then no solution

homework: done

>>8817476
I dont relly care what B equals, i just want it to stand alone on one side of the =

>>8817137
You don't without introducing extra variables.

>>8817476
What the fuck are you on about? It is always one by definition of sin and cos. Fucking leave this board, brainlet.

>>8817534
sin(x)^2+cos(x)^2=C

sin(x)^2 = 1/2 * (1-cos(2x))
cos(x)^2 = 1/2 * (1+cos(2x))


.'. sin(x)^2+cos(x)^2=C == 1/2 + 1/2 == 1

>>8817137
arccos((1-cos(2x))/2 + cos(x)^2) = arccos(C)
arccos(1/2) - x + x^2 = arccos(C)

x^2 - x + (Arccos(C+1/2)) = 0

solve quadratic and you get 2 values for x.

sin(x)^2+cos(x)^2=1 so C=1
from there you have sin(x)^2+cos(x)^2=1
or 1=1 from before
therefore x∈R

File: 352.png (1MB, 1920x1090px) Image search: [Google]

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>>8817139
Let [math]\mu[/math] be the Lebesgue measure. Then [math]\mu({1})=0[/math], so [math]C \neq 1[/math] almost everywhere.

>>8817664
Oops, [math]\mu(\{ 1\})=0[/math].

>>8817476
>actually, not very often
Actually fucking always brainletto