PreCal(Please check)
asked by
Abbey
Respond to this Question
Similar Questions

Math
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. 
PreCal(Please help)
Approximate the equation's soultions in the interval (o, 2pi). If possible find the exact solutions. sin 2x sinx = cosx I do not know where to start. 
PreCal
Approximate the equation's soultions in the interval (o, 2pi). If possible find the exact solutions. sin 2x sinx = cosx I do not know where to start. 
maths
hey, i would really appreciate some help solving for x when: sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x 2sinxcosx  cosx = 0 cosx(2sinx  1)=0 cosx = 0 or 2sinx=1, yielding sinx=1/2 from cosx=0 and by 
Trig
Verify the identity: tanx(cos2x) = sin2x  tanx Left Side = (sinx/cosx)(2cos^2 x 1) =sinx(2cos^2 x  1)/cosx Right Side = 2sinx cosx  sinx/cosx =(2sinxcos^2 x  sinx)/cosx =sinx(2cos^2 x 1)/cosx = L.S. Q.E.D. 
math
the problem is 2cos^2x + sinx1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is what i've done so far: 2cos^2x+sinx1=0 2cos^2x1+sinx=0 cos2x + sinx =0 1  2sin^2x + sinx = 0 
math
solve each equation for 0=/<x=/<2pi sin^2x + 5sinx + 6 = 0? how do i factor this and solve? 2sin^2 + sinx = 0 (2sinx  3)(sinx +2) sinx = 3/2, 2 how do I solve? These are not part of the special triangles. 2cos^2  7cosx + 
Trig.......
I need to prove that the following is true. Thanks (2tanx /1tan^x)+(1/2cos^2x1)= (cosx+sinx)/(cosx  sinx) and thanks ........... check your typing. I tried 30ยบ, the two sides are not equal, they differ by 1 oh , thank you Mr 
Math Help
Hello! Can someone please check and see if I did this right? Thanks! :) Directions: Find the exact solutions of the equation in the interval [0,2pi] cos2x+sinx=0 My answer: cos2x+sinx=cos^2xsin^2x+sinx =1sin^2xsin^2x+sinx 
Trig........
I need to prove that the following is true. Thanks (cosx / 1sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1sinx) multiply top