Ask a New Question
Search
Are these the solutions to sin2x sinx = cosx?
pi/2, 3pi/2, 3pi/4, pi/4, 7pi/4, 5pi/4
1 answer
Looks good to me.
Ask a New Question
or
answer this question
.
Similar Questions
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
Simplify #1:
cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer
1 answer
Simplify #1:
cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer
4 answers
1.Solve, finding all solutions in [0, 2π).
cosx sin2x + sinx cosx - sinx = 0 2.Solve, finding all solutions in [0, 2π). 3sec x
1 answer
more similar questions