2cosx+sinx+1=0 how do I get

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. 976 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. 910 views
Having trouble verifying this identity...cscxtanx + secx= 2cosx This is what I've been trying (1/sinx)(sinx/cosx) + secx = 2cosx
1. 2 answers
2. asked by Alex
3. 671 views
Differentiate(1- sinx) / (1 +sinx) How would I do this? The answer is (-2cosx)/ (1 + sinx)^2 Thanks in advance :)!!
1. 1 answer
2. asked by Allessandra
3. 509 views
Equations containing circular functionsand unit circle... solve 2cosx sinx + sinx = 0
1. 0 answers
2. asked by don
3. 467 views
Solving Equations Containing Circular Functions : Solve2cosx sinx + sinx = 0 when x is between 0 and 2pi
1. 1 answer
2. asked by don
3. 534 views
How do you simplify:(1/(sin^2x-cos^2x))-(2/cosx-sinx)? I tried factoring and creating a LCD of (sinx+cosx)(sinx cosx)
1. 2 answers
2. asked by Shannon
3. 583 views
by equating the coefficients of sin x and cos x , or otherwise, find constants A and B satisfying the identity.A(2sinx + cosx) +
1. 1 answer
2. asked by Candice
3. 643 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. 943 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. 936 views