SOLUTION: 2sin^2x-3sinx+1=0, 0 equal to or less than x <2pi

SOLUTION: 2sin^2x-3sinx+1=0, 0 equal to or less than x <2pi

Algebra.Com

Question 1007573: 2sin^2x-3sinx+1=0, 0 equal to or less than x <2pi
Answer by fractalier(6550) (Show Source): You can put this solution on YOUR website!
From 2sin^2x-3sinx+1=0, we factor and get
(2 sin x - 1)(sin x - 1) = 0 so that
sin x = 1/2 and sin x = 1 so that
x = pi/6 and 5pi/ 6 and pi/2
RELATED QUESTIONS
Solve each equation. Express each solution in degrees and in radians. 2cos squared... (answered by solver91311)

Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater (answered by Alan3354)

solve 2sinx-1= 0 for 0 is less than or equal to x less than 2pi (answered by god2012)

38) find all solution between 0 through 2pi 2sin^2x+... (answered by stanbon)

Find all the solutions to the equations in the interval (0, 2pi. 1. 2sin^2x=2+cosx 2.... (answered by lwsshak3)

how to solve... (answered by vasumathi)

2 sin^2x+sinx=1 Find all solutions to each equation for 0 less than or equal too x... (answered by stanbon)

Solve the trigonometric equation, using exact values when possible, where x is a real... (answered by jim_thompson5910)

Solve for all values of x if x is measured in degrees. 2cos^2x=3sinx 2(1-sin^2x)=3sinx (answered by stanbon)