Solve the equation 1 - 3cos^2 x = -5sinx for x in the range 0<(or equal) x <(or equal) to 2pi. Where x is correct to 3 decimal places.
thankkks !
1-3cos^2x = -5sin x
-3cos^x + 5sin x +1 = 0
-3(1-sin^2 x) + 5sin x +1 = 0
3sin^2 x + 5sin x -2 = 0
then x = sin x
replace sine with x so it will give you a quadratic equation.
3x^2 + 5 x -2=0
solutions are
x = -2/3 and x = -1
then your x was sine, so
sinx = -2/3 , sinx = -1
then if 0<(or =)x<(or=) 2pi
x=5.665 x = 4.712
first, cos^2x = 1- sin^2X.
then we get 1-3(1-sin^2x)=-5sinx
so, 3sin^2x+5sinx-2=0
(sinx+2)(3sinx-1)=0 --> sinx=-2 or sinx=1/3
however, sinx has to stay in range -1 to 1, therefore sinx can't be -2
then, sinx=1/3 -->using calculator we get, arcsin(1/3)=0.3398369095...
thus answer is 0.340 (to 3 decimal places)
and the another one is pi-0.3398369095=2.802 (in 4th quard)