SOLUTION: Sove for x when [0,2pi) [sin(2x) sin(x)]/2=cosx
Algebra.Com
Question 868663: Sove for x when [0,2pi) [sin(2x) sin(x)]/2=cosx
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Sove for x when [0,2pi) [sin(2x) sin(x)]/2=cosx
cancel out 2cos(x)
sin^2(x)=1
sin(x)=±1
x=π/2, 3π/2
RELATED QUESTIONS
Solve for 0≤x<2pi:
1-((sin^2x)/(1-cosx))=(1/2)
(Hint: simplify left side with (answered by Boreal)
I need to solve the following for x where degrees 0 (answered by Edwin McCravy)
Solve sin 2x = sin x for x between 0 and 2pi. (answered by stanbon)
Solve the equations of [0, 2pi]:
cosx(2sinx-1)=0
and... (answered by lwsshak3)
Find all solutions for, 5 sin^2(x) plus sin(x)= cos^2(x), that lie between 0 =< x =<... (answered by lwsshak3)
solve sin^2 x + (sinx)(cosx) =... (answered by Alan3354)
Solve on the interval [0, 2pi]
Sin*2x - sin x + 1=... (answered by lwsshak3)
Find all exact solutions on (0, 2pi)
sec(x) sin(x) - 2 sin(x) =... (answered by Boreal,MathLover1)
Solve the following for 0 < x < 2pi
(sin x + cos x)^2 =... (answered by lwsshak3,Alan3354)