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.
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
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.
the problem is 2cos^2x + sinx-1=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+sinx-1=0 2cos^2x-1+sinx=0 cos2x + sinx =0 1 - 2sin^2x + sinx = 0
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 +
I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr
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^2x-sin^2x+sinx =1-sin^2x-sin^2x+sinx
I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 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/(1-sinx) multiply top