/sci/ can you tell me where I am going wrong? What am I not seeing?

>>336976
sin^2 theta minus itself is zero not negative sin^2theta

>>336978
im getting:

1sin^2 x being subtracted by 2sin^2 x

1-2 is -1

1sin^2-2sin^2 = -1sin^2

>>336976
theta is quadratical to the minus, you have to use the feces theory that is derived from feceticus FUCK FUCK FUCK

>>336976
The first - sign on the right hand side came out of nowhere

>>336981
on the second line i mean

>>336982

my erasing was horrible

the equation is

sin^2theta = 4 - 2cos^2theta

Why did you square root both sides then divide by -1? You have a -sin^2(theta) if you take the square root of that you will have isin(theta) if you divide by negative one before square rooting you will get an i somewhere on the right instead. You also haven't solved for theta, you got part way to the solution and stopped.

no solutions, look at the problem again

If this doesn't work I will kill myself.
[math]sin^2(\Theta) = 4 + 2cos^2(\Theta) \\
sin^2(\Theta) = 4 + 2(1-sin^2(\Theta )), \left [sin^2(\alpha ) + cos^2(\alpha ) = 1\right ] \\
sin^2(\Theta) = 6 - 2sin^2(\Theta )\\
3sin^2(\Theta) = 6 \\
sin^2(\Theta ) = 2 \\
\Theta = arcsin(\sqrt2) + 2k\pi \\
\Theta = arcsin(-\sqrt2) + 2k\pi \\[/math]

>>336984

i stopped because to find theta i must take the cosine inverse of the square root of 2 which is not in the domain of cosine. theta must be equal to or greater than -1 and less than or equal to 1.

I just tried what you suggested that I dont think that is correct either, the solution must be a real number.

No solution, or the solution is i depending on what class you are in.
sin(x)^2 = 4 + 2cos(x)^2

1-cos(x)^2 = 4 + 2cos(x)^2

-cos(x)^2 = 3 + 2cos(x)^2

-3cos(x)^2 = 3

cos(x)^2 = -1

cos(x) = sqrt(-1) = i

>>336986
it looks fine for higher math, but its not a solution in lower math classes if a new variable is presented in the solution.

>>336986

I am sorry for the confusion, the right side of the equation should be 4-2cos^2theta.

My poor erasing made it seem like a plus.

>>336976
Is this a fucking joke? Summerfags, out.

>>336990
Well, if the right side did include the (+) side, you did good as far as I can see

>>336988

I am in trig at the moment, i thought it was no solution. Ill find out when I turn it in.

sin(theta)^2 = -2
sin(theta) = -sqrt(2)

WRONG did you pay attention in algebra?

x^2 = 2
x = +/-sqrt(2)

x^2 = -2
No solution. (you'll learn about imaginary numbers later)

>>336994

yes i have, ive learned imaginary numbers before, but i dont think it applies to trig, so i think this is just no solution

just need confirmation i guess

>>336993
circle no solution, but write a sentence explaining that since sqrt(-1) = i, that it could be (i) too. You can ALWAYS write comments in your answer to any question you are unsure about. At the very least, it saves your train of thought so if you get the question wrong, you can discuss the comment and answer with the professor.

>>336990
Pretty much everyone already answered but
[math]sin^2(\Theta) = 4 - 2cos^2(\Theta) \\
sin^2(\Theta) = 4 - 2(1-sin^2(\Theta)) \\
sin^2(\Theta) = 4 - 2 + 2sin^2(\Theta)\\
-sin^2(\Theta) = 2 \\
sin^2(\Theta) = -2\\
\Theta = arcsin(\sqrt-2) + 2k\pi\\
\Theta = arcsin(-\sqrt-2) + 2k\pi \\[/math]
Not pretty sure about the last thetas == but you shouldn't worry about them

>>336997
sin(theta) in trig is basically f(x) for instructional purposes, so saying it arcsin(+/-sqrt(2)i) + 2kx is redundant for anon.

If he used that answer, he'd likely be caught for cheating lol

Your handwriting is shit-tier, I cleaned it as much
as I could, but it is still shit. Work on that, your
future depends upon it.
max{cos} = 1, so min{right side} = 2
and max{left side} = 1, so no solution.
I hate these fckn trick questions.

File: 1498363202871.jpg (9KB, 500x125px) Image search: [Google]

9KB, 500x125px

>>336999
← neglected the image

>>337000
nice werke

>>336976
>not dotting all your 'i's

i posted this in the other thread
>>8995137
but it makes sense that it has to be complex because cos^2 and sin^2 are bounded between 1 and 0 with real solutions so the best you can hope for is 1=4-2 which cant happen unless the answer is a complex number

>>336976
use the power reducing formula

>>337003
>1 = 4 - 2
no

>>336995
-sin(x) = sin(-x)

>>336983
sin2 t = 4 - 2cos2 t
sin2 t + cos2 t + cos2 t = 4
but 1 + cos2 t < 2
fail

sin2 t = 4 + 2cos2 t
1 >= sin2 t = 4 - 2cos2 t >= 2
fail

>>337009
>1 >= sin2 t = 4 - 2cos2 t >= 2
it is 4 + 2cos2 t but doesn't change a thing