Hi, I' doing an additional maths past paper and don't know how to do this five mark question: Solve the equation 5sin2x=2cos2x in the interval 0<x<360 Give your answers correct to one decimal place.
(the more than and less than signs should be including 0 and 360) Thank you!
Hi, I' doing an additional maths past paper and don't know how to do this five mark question: Solve the equation 5sin2x=2cos2x in the interval 0<x<360 Give your answers correct to one decimal place.
(the more than and less than signs should be including 0 and 360) Thank you!
so you know the trigonometric identity : sinx / cosx = tanx ?
use this:
divide both sides by cos2x
so 5sin2x/cos2x = 2cos2x/cos2x 5tan2x = 2 tan2x = 0.4 2x = tan^-1(0.4) = 21.8 , 201.8, 381.8, 561.8(, 741.8) (degrees) (because it's 2x, limit is now 720 degrees) x = 10.9, 100.9, 190.9, 280.9(, 370.9) because range is from 0-360 degrees, x = 10.9, 100.9, 190.9, 280.9 (degrees)