SOLUTION: Solve the equation for the interval [0,2л):
tan2x-2cosx=0
PLEASE HELP ME!!!!!
Algebra.Com
Question 181755: Solve the equation for the interval [0,2л):
tan2x-2cosx=0
PLEASE HELP ME!!!!!
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
tan(2x)-2cos(x)=0
2(tanx/(1-tan^2(x))) - 2cosx = 0 (Identity from Google)
(tan(x))/(1-tan^2(x)) - cos(x) = 0
tan - cos(1-tan^2) = 0 All arguments x
tan - cosx + sin^2/cos = 0
sin - cos^2 + sin^2 = 0
sin + (cos^ + sin^2) = 2cos^2
sin + 1 = 2 - 2sin^2
2sin^2 + sin - 1 = 0 Quadratic in sin(x)
sin = (-1 +- sqrt(9))/4
sin(x) = 1/2, -1
x = pi/6, 3pi/2, 5pi/6
RELATED QUESTIONS
Solve each equation for the interval [0,2л):
A. (sin2x+cos2x)²=1
B. tan2x-2cosx=0
(answered by Alan3354)
Solve the equation for the interval [0,2л):
(sin2x+cos2x)²=1
PLEASE HELP... (answered by josmiceli)
Solve each equation graphically. Show all graphs.
A. 4sinx-cosx+2=0... (answered by Alan3354)
please help me solve this
1. prove the identity sinx/1-cosx =1/tanx
2. hence solve... (answered by Alan3354)
Solve the equation 2cosx + √3= 0 for 0 = x = 2π. (answered by Alan3354)
Please help me solve this equation for all the solutions in the interval [0, 2 pi)... (answered by stanbon)
Solve equation, sin(2x) - sinx - 2cosx + 1 = 0, on the interval 0 < or equal x <... (answered by Boreal)
Find all solutions of the equation in the interval
[0, 2 pi)
-sin2x+2cosx=0
(answered by ikleyn)
solve 2cos2x-2cosx=0 over the interval... (answered by MathLover1)