Respond to this Question
Similar Questions

Trig
prove the identity (sinX)^6 +(cosX)^6= 1  3(sinX)^2 (cosX)^2 sinX^6= sinx^2 ^3 = (1cosX^2)^3 = (12CosX^2 + cos^4) (1cosX^2) then multiply that out 12CosX^2 + cos^4  cosX^2 + 2cos^4 cos^6 add that on the left to the cos^6, and … 
Trig
Verify the identity: tanx(cos2x) = sin2x  tanx Left Side = (sinx/cosx)(2cos^2 x 1) =sinx(2cos^2 x  1)/cosx Right Side = 2sinx cosx  sinx/cosx =(2sinxcos^2 x  sinx)/cosx =sinx(2cos^2 x 1)/cosx = L.S. Q.E.D. 
trig
find the exact solutions 2cos^2x+3sinx=0 the way it stands, that is a "nasty" question. Are you sure the second term isn't 2sin(2x) ? 
math
the problem is 2cos^2x + sinx1=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+sinx1=0 2cos^2x1+sinx=0 cos2x + sinx =0 1  2sin^2x + sinx = 0 2sin^2x+sinx1=0 … 
Trigcheck answer
Solve the equation of the interval (0, 2pi) cosx=sinx I squared both sides to get :cos²x=sin²x Then using tri indentites I came up with cos²x=1cos²x Ended up with 2cos²x=1 Would the answer be cos²x=1/2? 
trig
how can i find all the exact solutions to the equation : 2cos^2x + 3sinx = 3 the solutions have to be between [0,2pi) 
Trig.
Can someone please help me with this? Find all solutions to the equation in the interval [0,2pi) cos2x=sinx I know I have to use some sort of identity, but I have no idea how to go about to solve this. 
PreCalculus
Find all solutions to the equation in the interval [0, 2pi) cos4xcos2x=0 So,this is what i've done so far: cos4xcos2x=0 cos2(2x)cos2x (2cos^2(2x)1)(2cos^2(x)1) No idea what to do next. 
Precalculus
Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π). 8. cos2x=cosx 10. 2cos^2x+cosx=cos2x Solve algebraically for exact solutions in the interval [0,2π). Use your grapher only to support your … 
Math Help
Hello! Can someone please check and see if I did this right?