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 !

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- November 3rd 2008, 01:02 PMJen1603Trig
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 ! - November 3rd 2008, 02:14 PMankitrig
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 - November 3rd 2008, 02:22 PMpnyfat
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) - November 3rd 2008, 02:24 PMpnyfat