SOLUTION: How do I solve
sin(x)-sin(2x)+2cos(x)-1=0
Algebra.Com
Question 633154: How do I solve
sin(x)-sin(2x)+2cos(x)-1=0
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ -\ \sin(2x)\ +\ 2\cos(x)\ -\ 1\ =\ 0)
Use
\ \equiv\ 2\sin(\varphi)\cos(\varphi))
To write:
\ -\ 2\sin(x)\cos(x)\ +\ 2\cos(x)\ -\ 1\ =\ 0)
Collect the 
terms in the LHS and the rest in the RHS:
\ -\ 2\sin(x)\cos(x)\ =\ 1\ -\ 2\cos(x))
Factor out )
\left(1\ -\ 2\cos(x)\right)\ =\ 1\ -\ 2\cos(x))
Multiply both sides by
})
\ =\ 1)
John
My calculator said it, I believe it, that settles it
RELATED QUESTIONS
Prove:
1-cot^2x/1+cot^2x=2cos^2-1
I tried this:
LHS... (answered by Alan3354)
show that... (answered by jim_thompson5910)
2cos^2(x/2)=(sin^2x)/(1-cosx) (answered by lwsshak3)
Prove that... (answered by ikleyn)
Prove that sin[2cos^-¹(x)]=(2x)√(1 -... (answered by ikleyn)
I need to solve the following for x where degrees 0 (answered by Edwin McCravy)
Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater (answered by Alan3354)
How do I get the left side to equal the right?
(sin x)(tan x cos x-cot x cos x) =... (answered by Tatiana_Stebko)
Why 1-sin^2x/sinx-cscx? is -sinx??? I thought it was sinx, but the answer says -sinx, how (answered by josgarithmetic,MathTherapy)
