SOLUTION: Solve the eq on the interval [0,2pi) : sin2x-cosx=0
I used the indetity sin2x = 2sinxcosx
So 2sinxcosx-cosx=0
Cosx(2sinx-1) = 0
Thus cosx=0 and 2sinx-1=0
Simplified i
Algebra.Com
Question 1121834: Solve the eq on the interval [0,2pi) : sin2x-cosx=0
I used the indetity sin2x = 2sinxcosx
So 2sinxcosx-cosx=0
Cosx(2sinx-1) = 0
Thus cosx=0 and 2sinx-1=0
Simplified im left with these to evaluate
Where is cosx = 0 and where is sinx= -1/2
Cosx = 0 at 3pi/2 and pi/2
Sinx = -1/2 at 7pi/ and 11pi/6
Would those be my values?
Found 2 solutions by solver91311, Alan3354:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ -\ 1\ =\ 0\ \text{ then } 2\sin(x)\ =\ 1\text{ and } \sin(x)\ =\ \frac{1}{2}\text{, not } -\frac{1}{2}.)
John
My calculator said it, I believe it, that settles it
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Solve the eq on the interval [0,2pi) : sin2x-cosx=0
------
Is it sin(2x)-cos(x) = 0 ?
Or sin^2(x) - cos(x) = 0 ?
---------------------
It's your job to make it clear.
RELATED QUESTIONS
I need help solving the following equation for 0(<=) X (<=)2pi
2sinXcosX= (sqrt... (answered by Fombitz)
sin2x - cosx + 1 =... (answered by fractalier)
Find all solutions in the interval [0, 2pi].... (answered by lwsshak3)
I need to verify that the equation is an identity.
{{{(sinx-cosx)^2=1-sin2x}}}
I... (answered by scott8148)
Calculating trigonometric equations for x in radians.
Find all the solutions of the... (answered by greenestamps)
Solve for (0<=x<=360)
sin2x-cosx=-1+2sinx... (answered by Alan3354,ikleyn)
Find all solutions to 3sin(2x) - cosx = 0 over the interval [0,360]. Round all... (answered by Alan3354)
Hello (again),
Please disregard the first question sent to you from... (answered by vleith)
sin2x=2sinxcosx (answered by roseo)