Saturday
August 23, 2014

Homework Help: math

Posted by alex on Monday, April 7, 2008 at 9:02am.

the problem is
2cos^2x + sinx-1=0

the directions are to "use an identity to solve each equation on the interval [0,2pi).

This is what i've done so far:

2cos^2x+sinx-1=0

2cos^2x-1+sinx=0

cos2x + sinx =0

1 - 2sin^2x + sinx = 0

-2sin^2x+sinx-1=0

2sin^2x-sinx+1=0

I think i'm supposed to factor it next but i'm not sure?

(2sinx-?)(sinx-?)

would doing that step be correct?

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Trig - Verify the identity: tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(...
math - solve each equation for 0=/<x=/<2pi sin^2x + 5sinx + 6 = 0? how do ...
Trigonometry - Find all solutions on the interval (0,2pi): 2-2cos^2=sinx+1
trig - find the exact solutions 2cos^2x+3sinx=0 the way it stands, that is a "...
Trig Help - Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx...
Math (trigonometry) - Trigonometry identities are so hard... I need some help ...
math - tanx+secx=2cosx (sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx ...
Math-gr12 - How would I solve this equation for x in the interval 0<x<2pi ...
Trig - prove the identity (sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= ...
Trig--check answer - Solve the equation of the interval (0, 2pi) cosx=sinx I ...

Search
Members