Posted by **Hannah** on Monday, March 1, 2010 at 4:16pm.

Approximate the equation's solutions in the interval (0,2pi).

sin2x sinx = cosx

I know that I should work on the left side first. I know how to solve for the intervals but I am just not sure how to start this off.

- Math -
**Reiny**, Monday, March 1, 2010 at 5:17pm
sin2x sinx = cosx

2sinxcosxsinx - cosx = 0

cosx(2sin^2x - 1) = 0

cosx = 0 or sin^2x = 1/2

if cosx = 0, x = pi/2 or 3pi/2

if sin^2x = 1/2

sinx = ± 1/√2

(x could be in all 4 quadrants)

x = pi/4, pi-pi/4, pi+pi/4, and 2pi-pi/4

= ....

(I don't understand why they want you to approximate, we have exact answers)

## Answer This Question

## Related Questions

- check math - Here are my answers. Can you check if I got the right answers? ...
- Trig - Verify the identity: tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(...
- Pre-Cal(Please check) - Approximate the equation's solutions in the interval (0,...
- Pre-Cal - Approximate the equation's soultions in the interval (o, 2pi). If ...
- Pre-Calc - Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= ...
- Pre-Cal(Please help) - Approximate the equation's soultions in the interval (o...
- maths - hey, i would really appreciate some help solving for x when: sin2x=cosx ...
- math - solve each equation for 0=/<x=/<2pi sin^2x + 5sinx + 6 = 0? how do ...
- Math - Im really struggling with these proving identities problems can somebody ...
- Precalculus - Please help!!!!!!!!!!! Find all solutions to the equation in the ...

More Related Questions