SOLUTION: 2sin^2x=3sinx+5
solve equation on interval [0,2pi)
Algebra.Com
Question 1001690: 2sin^2x=3sinx+5
solve equation on interval [0,2pi)
Answer by ikleyn(52847) (Show Source): You can put this solution on YOUR website!
.
= + .
Let us introduce a new variable, y = sin(x).
Then the equation takes the form
= , or
- - = .
Solve it by using the quadratic formula:
= = = .
= -1 -----> sin(x) = -1 -----> x = .
= doesn't fit, because sin(x) can not be more than 1.
Answer. x = .
RELATED QUESTIONS
Find all the solutions to the equations in the interval (0, 2pi.
1. 2sin^2x=2+cosx
2.... (answered by lwsshak3)
solve 2sin^2x+3sinx-4=0 in the interval... (answered by Alan3354)
Solve for theta on the interval 0 <= theta <= 2pi.
2sin(theta)cos(theta) - cos(theta)... (answered by Fombitz)
graph y= 2sin(2x - pi) on the interval 0 <= x <=... (answered by nyc_function)
Solve the equation on the interval [0,2pi)
2sin^2x=-3sinx+5, in terms of pi and use... (answered by lwsshak3)
Solve each equation for 0 (answered by stanbon)
Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater (answered by Alan3354)
Solve equation, sin(2x) - sinx - 2cosx + 1 = 0, on the interval 0 < or equal x <... (answered by Boreal)
38) find all solution between 0 through 2pi
2sin^2x+... (answered by stanbon)