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

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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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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.
 
[tex]\tan^2 x -\sqrt 3 \tan x=0[/tex]

Correct?
 
Last edited:
sorry let me try again..


tan^2x - (square root of 3) *tanx=0
 
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.
 
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
 
How did you simplify it to sinx=1/2?
 
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.
 
[tex]\tan 2x \cot x -3 =0[/tex]

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.
 
ok thanks,

Another problem, same directions:

sin2x=2cosx

attempt:

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

stuck on what to do from here
 
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.
 
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