Simplify: (cosx + cos2x +

Solve the equation of the interval (0, 2pi)cosx=sinx I squared both sides to get :cos²x=sin²x Then using tri indentites I came
1. 4 answers
2. asked by Brian
3. 821 views
Simplify #1:cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer
1. 4 answers
2. asked by Anonymous
3. 1,976 views
Simplify #1:cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer
1. 1 answer
2. asked by Anonymous
3. 1,381 views
Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π).8. cos2x=cosx 10. 2cos^2x+cosx=cos2x Solve
1. 5 answers
2. asked by Anonymous
3. 869 views
Solve:cos(2x-180) - sin(x-90)=0 my work: cos2xcos180 + sin2xsin180= sinxcos90 - sin90cosx -cos2x - sin2x= cosx -cos^2x + sin^2x
1. 0 answers
2. asked by Jess
3. 801 views
Having trouble with this problem:Rewrite the expression as an equivalent expression that does not contain powers greater than 1.
1. 2 answers
2. asked by Brian
3. 1,476 views
Prove that cos3x/cosx-cos6x/cos2x=2(cos2x-cos4x)
1. 2 answers
2. asked by Joel
3. 1,187 views
Where do I start to prove this identity:sinx/cosx= 1-cos2x/sin2x please help!! Hint: Fractions are evil. Get rid of them. Well,
1. 0 answers
2. asked by maria
3. 847 views
Simplify the expression using trig identities:1. (sin4x - cos4x)/(sin2x -cos2x) 2. (sinx(cotx)+cosx)/(2cotx)
1. 1 answer
2. asked by Tasha
3. 2,891 views
solve algebreically the equationcos2x=1-sinx also cosx=cosx-1 and please also sin2x=3sinx
1. 1 answer
2. asked by adil ahmed
3. 665 views