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Can someone help me solve this? cos(x^2+45)+tan(4x+90) = sin(3x+120) A step by step would be great.

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by solving do you mean proving?
Yes. :)
What methods have been discussed?

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Other answers:

We are solving for x not proving this. This isn't an identity.
Or does it say to disprove or prove?
Ah yes, my bad. It says solve.
Can you tell me what methods in class have been discussed?
None, this is from a book with only problems I picked up at barnes and noble. We haven't done this and wont do it for another year or two. I've had no problems with the book up to this point.
can you tell me what methods the book has discussed?
None. It's a book of only problems. It starts from Geometry and works it's way up in problems.
Name of book please
Math Review for the High School Classroom
I looked for it online and can't find it.
I didn't see the book. :( So there are no examples?
No examples, just questions. :c How did you learn it?
Honestly, I don't know without seeing some example or them mentioning some method they want to use.
Ah, okay. Thanks, anyway. :)
cos(x^2+45)+tan(4x+90) = sin(3x+120) cos(x^2+45)+sin/cos(4x+90) = sin(3x+120) dividing everything by sin. tan(x^2+45) + sec(4x+90) = 3x + 120 i don't remember trig that well, but you can convert it to this i think.
You can't do that... Like you can't divide by the trig part The trig part is a function not a number or variable on its own
Are the angles in radians or degrees?
I'll assume degrees for now.
Degrees, yeah. ^^
45,90,120 suggest they are in degrees... we can convert sin and cos in tan, like cos A = tan (sqrt (1-A^2)/A) something like this and then use tan A+tan B formula...... but if thats possible, the algebra is going to be very ugly....
but @hartnn tan(90) does not exist you were talking about the sum identity for tan right?
@myininaya tan(a+90) = -1/tan(a), since they're perpendicular.
not tan(A+B) tan A +tan B , where B = 4x+90 and yeah, tan (A+90) = - cot 90 so maybe convert cos into cot and use cotA-cotB formula... i don't think that will be of any use though....
I don't like the x squared part much.
Series solution is what I'd prefer to us,e but not without an attempt with standard algera.
or maybe the whole problem i don't see how to do it with algebra/trig He said it was for high school though
even wolf doesn't give integral or rational solutions...
Yes, I'm a junior in high school, but the book said High School. It extends way past what I expect to learn in High School, though. I think they may have the age range a bit messed up.
No simple way to solve at that level.
@primeralph if you want, can you show me your series thing you were talking about? You know if you want to.
I might just go and ask one of the professors at the college I intern at. Thanks for your help, guys. :)
First I'd have to convert all to radians.
Then find a way to represent the equation as a DE, or simply use the Taylor power series.
what is equivalent to a (1 radian) squared? That bugs me. The little square there.
What I'm saying is I'm not used to looking at angles being squared
@myininaya Yeah I understand, but you'd have to extend your thinking beyond expecting trig to only apply to angles.
It will be extensive, but in the end, there might be real solutions.
x^2 and 45 will have different units
we have x^2=a^2 (degs)^2 and 45 is just in degs not degs^2
The x^2 term will be set to have some correction factor. After taking a look at it, I think it's solvable algebraically. It's just long.
And with a lot of restrictions to be imposed.
those ugly solutions
Don't think you used radians. Wolfram is radians by default.
doesn't make much difference
wolfram doesn't want to solve it this is the way input it cos((x deg)^2+45 deg)+tan(4x deg+90 deg)=sin(3x deg+120 deg)
Maybe it doesn't like the unlike units being added either idk

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