Posted by Natash on Saturday, November 22, 2008 at 9:08pm.
Find all solutions of the equation 2sin^2x-cosx=1 in the interval [0,2pi)
- trig - Damon, Saturday, November 22, 2008 at 9:33pm
2 (1 - cos^2 x) - cos x = 1
2 - 2 cos^2 x -cos x = 1
2 cos^2 x + cos x -1 = 0
let y = cos x
2 y^2 +y -1 = 0
(y+1)(2y-1) = 0
y = .5 or y = -1
cos x = .5 or cos x = -1
x = 60 degrees or 300 degrees
or 180 degrees
- trig - drwls, Saturday, November 22, 2008 at 9:37pm
2(1-cos^2x)-cosx = 1.
2cos^2x + cos x -1 = 0,
which can be factored to provide:
(2cosx -1)(cosx +1) = 0.
cosx = 1/2 or -1
Use that to find the three angles.
Answer this Question
- Math - Can I please get some help on these questions: 1. How many solutions does...
- Math Trig - Find all solutions on the interval [0,2pi] for the following: 2sin^2...
- trig - I need to find all solutions of the given equations for the indicated ...
- calculus - Find all solutions to the equation in the interval [0,2pi) Cosx-cos2x
- Math - Directions: Find all solutions of the equation in the interval (0, 2pi) ...
- Trigonometry - Find all solutions of the equation in the interval [0,2pi] ...
- trig - Find ALL solutions to the equation on the interval [0, 2pie]: 2sin(1/2X...
- Trig - Find all solutions of the equation on the interval [0,2pi): Tan^2x=1-secx
- Trig - Find all solutions of equation on interval [0,2pi] 1=cot^2x + cscx
- Trig - Find all of the solutions between 0 and 2pi: 2sin(x)^2 = 2 + cos(x)