Sin2xcosx Cos2xsinx - How To Discuss
Sin2xcosx Cos2xsinx
Solve this equation? sin2xcosx = cos2xsinx? 3
The correct answer is x = n (pi), n = 0, + / 1, + / 2 ....
I need to use the addition and subtraction formula.
I have created sin2xcosx cos2xsinx = 0.
Sin (2xx) = 0
So I st. Help?
Sin 2x cos x cos 2x sin x = 0
Sin (2x x) = 0
Sin = 0
x = n € for n = 0, 1, ± 2, 3,
Problem 1: pi / 2.3pi / 2.5pi / 2 ......> cos2x + sinx = one million one million2sinx x + sinx = 0 so take (one million one) sinx = update your square equation in t Then use (one million) and update to X. Issue 2: sin2xcosxsinx = 0 2sinxcosxsinx = 0 sinx (2cos xa million) = 0 sinxcos2x = 0 then both (one million) sinx = 0 or (2) cos2x = 0 then for each and the total cost of pi x = 0 , 2pi ... and relax with (2) x = pi / 4 .. :)
