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Really Hard Trig Equation Someone help please?

  • #1

Homework Statement


Solve the equation:
cos2x - sinx = (1/2) for [0,2pi).


Homework Equations


sin^2x + cos^2 = 1


The Attempt at a Solution



cos^2(2x) - sin^2x = 1
1 = 1 Infinite Solutions?
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
4,302
47
[tex](\cos 2x - \sin x)^2 = \cos^2(2x) - 2 \cos(2x) \sin(x) + \sin^2(x) \neq \cos^2(2x) - \sin^2(x)[/tex]
and
[tex]\cos^2(2x) - \sin^2(x) \neq \cos^2(x) + \sin^2(x)[/tex]
Maybe
[tex]\cos(2x) = \cos^2(x) - \sin^2(x)[/tex]
will help?
 
  • #3
63
0
Please rewrite the question. There is repeated misuse of exponents and misplaced variables. Then I'll help all I can!
 
  • #4
cos2x - sinx = (1/2) for [0,2pi).

How does that have misplaced exponents and variables? Would Writing parentheses help? Im really sorry for the confusion and my mistake.

Cos(2x) - sin(x) = (1/2) for [0,2pi).

Cosine of 2x minus sin of x equals 1 half for the range 0 through 2pie
 
  • #5
1,752
1
Please rewrite the question. There is repeated misuse of exponents and misplaced variables. Then I'll help all I can!
slap yourself.

Ok, rather than the identity your using ... use:

[tex]\cos{2x}=1-2\sin^2 x[/tex]

Just factor and it's solved!!!
 
Last edited:
  • #6
So then

1 - 2sin^2(x) - sin(x) = (1/2) [-(1/2)]

-2sin^2(x) - sin(x) + (1/2) = 0

Quadratic Formula?

I tried it and got sinx = (1 +/- Sqrroot(5)) / -4)
x = -54 and 18.
Somehow that doesn't seem right because radians aren't that big on problems. Any help?
 
  • #7
63
0
-54 and 18 are the answers in degrees, and they are correct
 
  • #8
brendanH the question says between 0 and pi. therefore even if you are right -54 doesnt work
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
41,794
925
And, indeed, since the question itself uses [itex]\pi[/itex], the answer should be in radians, not degrees. There are two solutions between 0 and [itex]\pi[/itex].
 

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