Ask a New Question
Search
Given the equation 2cos^2x+sinx-1=0 1.
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
3 answers
asked by
alex
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
0 answers
asked by
Ashley
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 =
0 answers
asked by
shan
979 views
solve each equation for 0=/<x=/<2pi
sin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx - 3)(sinx
7 answers
asked by
sh
948 views
(cosx - sinx)^2 + (cosx + sinx)^2 = 2
Iam in this step: 2cos^2(x) + 2sin^2(x) = 2 How can I make the equation on the right equal
2 answers
asked by
Luis
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)]
3 answers
asked by
Anonymous
944 views
Find the derivative
y=sinx(sinx+cosx) I got y'= cos(x)-sin^2(x)+2cos(x)sin(x) is this righttttttt??
4 answers
asked by
Noe
609 views
Find the solutions to the equation cos2x=sinx and 2cos^2x-3cosx-3=0
0 answers
asked by
Jill
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 answer
asked by
Anonymous
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)
0 answers
asked by
JungJung
913 views