Given the equation 2cos^2x+sinx-1=0 1.

the problem is2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
1. 3 answers
2. asked by alex
3. 866 views
Verify the identity:tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx
1. 0 answers
2. asked by Ashley
3. 1,247 views
tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
1. 0 answers
2. asked by shan
3. 979 views
solve each equation for 0=/<x=/<2pisin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx - 3)(sinx
1. 7 answers
2. asked by sh
3. 948 views
(cosx - sinx)^2 + (cosx + sinx)^2 = 2Iam in this step: 2cos^2(x) + 2sin^2(x) = 2 How can I make the equation on the right equal
1. 2 answers
2. asked by Luis
3. 555 views
Prove the following:[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
1. 3 answers
2. asked by Anonymous
3. 944 views
Find the derivativey=sinx(sinx+cosx) I got y'= cos(x)-sin^2(x)+2cos(x)sin(x) is this righttttttt??
1. 4 answers
2. asked by Noe
3. 609 views
Find the solutions to the equation cos2x=sinx and 2cos^2x-3cosx-3=0
1. 0 answers
2. asked by Jill
3. 530 views
Prove:sin^2x - sin^4x = cos^2x - cos^4x What I have, LS = (sinx - sin^2x) (sinx + sin^2x) = (sinx - 1 -cos^2x) (sinx + 1 -
1. 1 answer
2. asked by Anonymous
3. 540 views
prove the identity(sinX)^6 +(cosX)^6= 1 - 3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1-cosX^2)^3 = (1-2CosX^2 + cos^4) (1-cosX^2)
1. 0 answers
2. asked by JungJung
3. 913 views