Posted by **greg** on Monday, December 3, 2012 at 10:19am.

Solve sin2x + sin4x = cos2x + cos 4x for x in the interval[0,2pi)

Hint: the following substitution should come in handy: sin3x= cos3x . tan3x

## Answer This Question

## Related Questions

- Trigonometry - Verify the identities algebraically. 1.) TAN^5 X= TAN³X SEC²X-TAN...
- Math (Trig) - Rewrite the following expression in terms of tan3x: (sin2x + sin4x...
- Calculus - In the interval (0 is less than or equal to x which is less than or ...
- Trig - Simplify: (cosx + cos2x + cos3x + cos4x + cos5x + cos6x) / (sinx + sin2x...
- Trig - help please!: (sinx + sin2x +sin3x) / (cosx + cos2x + cos3x) = tan2x
- calc - Where do I start to prove this identity: sinx/cosx= 1-cos2x/sin2x please ...
- Trig--check answer - Solve the equation of the interval (0, 2pi) cosx=sinx I ...
- Calculus - Solve identity, (1-sin2x)/cos2x = cos2x/(1+sin2x) I tried starting ...
- calculus - Solve for x on the interval (0, 2pi] (sin2x + cos2x)^2 = 1
- Trig - Prove the following functions: (sinx+sin2x)/(1+cosx+cos2x)=tan x (cos3x/...

More Related Questions