SOLUTION: 2 sin^2x+sinx=1
Find all solutions to each equation for 0 less than or equal too x which is less than or equal to 2pi
Algebra ->
Trigonometry-basics
-> SOLUTION: 2 sin^2x+sinx=1
Find all solutions to each equation for 0 less than or equal too x which is less than or equal to 2pi
Log On
Question 892171: 2 sin^2x+sinx=1
Find all solutions to each equation for 0 less than or equal too x which is less than or equal to 2pi Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 2 sin^2x+sinx=1
Find all solutions to each equation for 0<= x <=2pi
---------------
2sin^2(x) + sin(x) -1 = 0
------
Use the quadratic formula to get:
sin(x) = [-1 +- sqrt(1-4*2*-1)]/4
----
sin(x) = [-1 +- sqrt(9)]/4
----
sin(x) = [-1 +- 3]/4
---
sin(x) = 1/2 or sin(x) = -1
-----
If sin(x) = 1/2, x = pi/6 or (5/6)pi
If sin(x) = -1, x = (3/2)pi
================================
Cheers,
Stan H.
-----------------------
(