Solve: #5sin theta = 6 cos^2theta -6# for #theta in [0, 2pi]#?

2 Answers
Jan 4, 2018

#theta = 0, pi, approx 4.127, approx 5.298,2pi # for #theta in [0,2pi]#

Explanation:

#5sin theta = 6 cos^2theta -6#

#5sin theta = 6(cos^2theta -1)#

#5sin theta = -6sin^2theta#

#sin theta(5+6sin theta) =0#

#sin theta = 0 or sin theta = -5/6#

#sin theta =0 ->theta = 0, pi or 2pi in [0, 2pi]#

#sin theta = -5/6 -> theta = arcsin (-5/6) approx 4.127 or approx 5.298 in [0, 2pi]#

Hence, #theta = 0, pi, approx 4.127, approx 5.298,2pi # for #theta in [0,2pi]#

We can see these results from the zeros of the graph of #sin theta(5+6sin theta)# for # theta in [0,2pi]# below:

graph{sinx(5+6sinx) [-1.56, 7.21, -1.892, 2.49]}

Jan 4, 2018

#theta=0; pi; 4.126; 5.298; 2pi#

Explanation:

#5sintheta=6cos^2theta-6#

#5sintheta=6(cos^2theta-1)#


#1=sin^2x+cos^2xquad=>quadsin^2x=1-cos^2x#


#5sintheta=-6(1-cos^2theta)#

#5sintheta=-6sin^2theta#

#5sintheta+6sin^2theta=0#

#sintheta(5+6sintheta)=0#

#sintheta=0quad=>quadtheta_1=0quad^^quadtheta_2=piquad^^quadtheta_3=2pi#


#5+6sintheta_(4,5)=0#
#=>sintheta_(4,5)=-5/6#
To find that you have to use calculator but it will tell you only one value. To find second value we need to know in which quadrant we are.

#=>theta_(4)~~-0.985~~-pi/3.1#
but #0<=theta<=2pi#
#=>theta_4~~-0.985+2pi~~color(pink)5.298~~(5pi)/3#

#(3pi)/2<=(5pi)/3<=2piquad=>quad# 4th quadrant
That means that the last angle must be in 3rd quadrant. To calculate that we use well known trick:
For 3rd quadrant: #varphi=pi+varphi_0#

Applying to our case: #theta_0=|-0.985|#
#=>theta_5=pi+0.985=color(orange)4.126#

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