solve 2cos^2x-sinx=1, 0 < or

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. 877 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. 959 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. 1,261 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. 992 views
27. Solve for all radian solutions. sinx = −2cos x
1. 1 answer
2. asked by Mark
3. 447 views
solve 2cos^2x-sinx=1, 0 < or equal to x < 2pie
1. 4 answers
2. asked by Calvin
3. 1,364 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. 950 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. 615 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. 548 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. 923 views