Pre Calc

posted by .

Solve the equation on the interval [0,2pi).
cos2x=(*2/2)

Similar Questions

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=1-cos²x Ended up with 2cos²x=1 Would the answer be cos²x=1/2?
2. Pre Calc

Solve the equation on the interval [0,2pi). cos2x=(*2/2)

find the solutions of the equation that are in the interval [0,2pi) 2-cos^2x=4sin^2(1/2x)
4. pre calc

cos2x=cosx in interval of [0,2pie]
5. Pre-Calculus

Find all solutions to the equation in the interval [0, 2pi) cos4x-cos2x=0 So,this is what i've done so far: cos4x-cos2x=0 cos2(2x)-cos2x (2cos^2(2x)-1)-(2cos^2(x)-1) No idea what to do next.
6. calculus

Solve for x on the interval (0, 2pi] (sin2x + cos2x)^2 = 1
7. Pre-Calc/Math

Given cos x=2/3 and 3pi/2<x<2pi, find the exact value of cos2x?
8. Pre-Calc/Math

Use factoring, the quadratic formula, or identities to solve cos(x)+1=sin^(2)x. Find all solutions on the interval [0, 2pi)
9. 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 …
10. Math

Can I please get some help on these questions: 1. How many solutions does the equation,2sin^2 2 θ = sin2θ have on the interval [0, 2pi]?

More Similar Questions