How Do You Solve Trigonometric Equations for Angles Between 0 and 180 Degrees?

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Homework Help Overview

The discussion revolves around solving trigonometric equations for angles between 0 and 180 degrees. The original poster presents several equations, including sin and cos functions, and expresses difficulty in understanding the concepts and methods involved in solving them.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore various identities and algebraic manipulations related to sine and cosine functions. There are attempts to factor equations and apply the quadratic formula. Questions arise regarding the validity of certain identities and the steps taken in the problem-solving process.

Discussion Status

The conversation is ongoing, with participants actively engaging in problem-solving together. Some guidance has been offered regarding algebraic manipulation and the use of identities, but there is no explicit consensus on all approaches or solutions yet.

Contextual Notes

Participants are working under the constraints of a homework assignment, with a focus on understanding rather than providing complete solutions. There is mention of an upcoming test, which adds urgency to the discussion.

  • #31
babacanoosh said:
ok thanks for your help,

tanx=0

x=180k degrees,
so the answer would be 360k degrees.
Yes, but don't forget your solutions are from 0-180.

tanx=0 whenever sinx=0.
 
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  • #32
ok, thanks
 
  • #33
i have a few new problems that i can't seem to get:

Solve 0-180 degrees

tan^2x- (square root of 3)tanx=0

Attempt:
simplified it all the way down to 1-ot^2x=sec^2x...stuck there, or i just did i way wrong

Next problem:

tan2xcotx-3=0

attempt:

simplified it to sinx=1/2...

i did it wrong, because this is the wrong answer.
 
  • #34
\tan^2 x -\sqrt 3 \tan x=0

Correct?
 
Last edited:
  • #35
sorry let me try again..


tan^2x - (square root of 3) *tanx=0
 
  • #36
What common term do you have that you could factor ... ? You explored this method with the other problems, refer back to them if you need.
 
  • #37
tangent...
tan(tan-square root of 3)=0
tan = 0

0 degrees and 180 degrees

tan = square root of 3

60 degrees,

so answer is 0, 60, 180 degrees

thanks, i don't understand why i didnt see that one, however the next one i am still stuck
 
  • #38
How did you simplify it to sinx=1/2?
 
  • #39
i don't even know anymore, it involved squaring the equation and then changing cot^2x into 1-csc^2x...im way off i guess.
 
  • #40
\tan 2x \cot x -3 =0

Use an identity to re-write tan2x then use another identity to get it in terms of x, not 2x, and change cotx in terms of sines and cosines. Cancel like terms and multiply like terms, and it's solved.
 
  • #41
ok thanks,

Another problem, same directions:

sin2x=2cosx

attempt:

2sinxcosx-2cosx=0
2sinx-cosx=0

stuck on what to do from here
 
  • #42
2sinxcosx-2cosx=0
2sinx-cosx=0

What was your reasoning for the 2nd step? Refer back to the 1st problem we worked on tonight ... I don't think you're thinking hard enough.
 
  • #43
I'm sorry, that's wrong,

well cos(2sin -2)=0
cos=0

90 degrees

sinx=1

90 degres

the final answer would be 90 degrees
 
  • #44
That's right.
 

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