SOLUTION: Solve the equation for x in the interval 0≤x ≤2 π:
2sin²x-sinx-1=0
Algebra.Com
Question 170922: Solve the equation for x in the interval 0≤x ≤2 π:
2sin²x-sinx-1=0
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve the equation for x in the interval 0≤x ≤2 π:
2sin²x-sinx-1=0
Factor to get:
(2sinx+1)(sinx-1) = 0
sinx = -1/2 or sinx = 1
x = (11/6)pi or (7/6)pi or pi/4 or (3/4)pi
==============================================
Cheers,
Stan H.
RELATED QUESTIONS
Solve the equation for all solutions in the interval 0degree ≤ x < 360degree.... (answered by Edwin McCravy)
Solve this trig equation in the interval 0≤x≤2π:
2/sinx+10=6
(answered by Alan3354)
Solve for x in the equation (1 +sinx/cosx)+(cosx/1+sinx)=4 for 0≤ x ≤ 2π (answered by HyperBrain)
Solve the following equation for x in the interval 0 ≤ x ≤ 2π
tan(x)... (answered by Edwin McCravy)
Solve the equation algebraically for "x" where 0≤x<2π.
sinx + cosx = 1... (answered by Fombitz)
Solve each equation for x in the interval 0≤x ≤2 π:
1/1+tan²x=-cosx
(answered by stanbon)
How many points in the interval 0 ≤ x ≤ 2π satisfy the equation cos 2x = (answered by stanbon)
How many points in the interval 0 ≤ x ≤ 2π satisfy the equation cos 2x = (answered by stanbon)
Solve the cos x-1 =0 on the interval 0≤x≤pi/2
x+ π/3
x=0
x=π/6... (answered by richard1234)