Math - Solving Trig Equations

What am I doing wrong?

Equation:
sin2x = 2cos2x

Answers:
90 and 270

....

My Work:

2sin(x)cos(x) = 2cos(2x)
sin(x) cos(x) = cos(2x)
sin(x) cos(x) = 2cos^2(x) - 1
cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1
cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1
5cos^4(x) - 5cos^2(x) + 1 = 0
cos(x) = (+/-) 0.85065080835215
cos(x) = (+/-)0.52573111211905

asked by Anonymous
  1. wouldn't it be easier to solve for 2x then divide it by two?

    sin2x=2cos2x

    sin2x/cos2x=2
    tan 2x=2

    2x=63.43 deg
    x= .... you do it.

    posted by bobpursley

Respond to this Question

First Name

Your Response

Similar Questions

  1. Math - Solving for Trig Equations

    Solve the following equation for 0 less than and/or equal to "x" less than and/or equal to 360 -- cos^2x - 1 = sin^2x -- Attempt: cos^2x - 1 - sin^2x = 0 cos^2x - 1 - (1 - cos^2x) = 0 cos^2x - 1 - 1 + cos^2x = 0 2cos^2x - 2 = 0
  2. TRIG!

    Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x +
  3. Calculus

    Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2,
  4. tigonometry

    expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b)
  5. Math - Solving Trig Equations

    Solve each equation for o is less than and/or equal to theta is less than and/or equal to 360 -- sin^2x = 1 = cos^2x -- Work: cos^2x - cos^2x = 0 0 = 0 -- Textbook Answers: 90 and 270 -- Btw, how would you isolate for cos^2x = 0?
  6. Precal

    I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = -
  7. trigonometry (please double check this)

    Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. 1. sin2Į = (sqrt 3)/2 2. sin^2Į =
  8. trig

    I don't understand this. can you please tell how you got each answer? describe how the graph of each function is related to a basic trigonometric graph. 1. y= 1/2cos x 2. y= 2cos x+1 3. y= sin x-4 4. y= sin(x-4) 5. y= sin(3x) 6.
  9. another please help me check~calculus maths

    y=3e^(2x)cos(2x-3) verify that d^2y/dx^2-4dy/dx+8y=0 plz help me i tried all i could but it become too complicated for me here set u=3e^(2x) v=cos(2x-3) du/dx=6e^(2x) i used chain rule dv/dx=-2sin(2x-3)
  10. math (trig)

    i have some problems doing trig the first one is: Show that cos(x/2) sin(3x/2) = ½(sinx + sin2x) i know that you are supposed to substitute all those trig function things in it but i kind of forgot how to the only that i can see

More Similar Questions