How do you use the quadratic formula to solve #12sin^2x-13sinx+3=0# in the interval #[0,2pi)#?

2 Answers
May 26, 2017

The solutions are #S={0.34,0.848,2.294,2.802}rd#

Explanation:

Our equation is

#12sin^2x-13sinx+3=0#

The quadratic equation is

#ax^2+bx+c=0#

We start by calculating the discriminant

#Delta=b^2-4ac=(-13)^2-4*(12)*(3)#

#=169-144=25#

As,

#Delta>0#, there are 2 real solutions

So,

The solutions are

#x=(-b+-sqrtDelta)/(2a)#

#sinx=((13)+-5)/24#

#sinx=18/24=3/4# or #sinx=8/24=1/3#

The values of #x# are

#x=arcsin(3/4)# and #arcsin(1/3)#

#x=0.848rd#, #x=2.294rd#, #x=0.34rd# and #x=2.802rd#

May 26, 2017

#x~~0.3398#, #x~~0.8481#, #x=2.294#, and #x=2.802# radians.

Explanation:

Strategy: Rewrite this equation as a quadratic equation using #u=sin(x)#. Solve the quadratic equation for #u# by factoring. Then replace #u# with #sin(x)# again and solve using #arcsin#.

Step 1. Rewrite this equation as a quadratic using #color(red)u=color(red)sin(x)#

You are given
#12(color(red)(sin(x)))^2-13(color(red)(sin(x)))+3=0#

Replace #color(red)(sin(x))# with #color(red)(u)#
#12color(red)(u)^2-13color(red)(u)+3=0#

Step 2. Factor the quadratic equation.
#(3u-1)(4u-3)=0#

Solving gives us
#3u-1=0# and #4u-3=0#
#u=1/3# and #u=3/4#

Step 3. Replace #u# with #sin(x)# again and solve with #arcsin#
#sin(x)=1/3# and #sin(x)=3/4#

#sin^(-1)(sin(x))=sin^(-1)(1/3)# and #sin^(-1)(sin(x))=sin^(-1)(3/4)#

#x~~0.3398# radians and #x~~0.8481# radians

These answer work because we were asked to find the solutions in #[0,2pi]~~[0,6.2832]#

However, the graph of #y=12(sin(x))^2-13(sin(x))+3# is

Desmos.com and MS Paint

Which shows four solutions in the interval #[0,2pi]#, not two.

We must find the other solutions by recognizing that sine functions are periodic with respect to #pi#

So, #x=pi-0.8481=2.294# and #x=pi-0.3398=2.802#

The other two solutions are
#x=2.294# and #x=2.802#