Can you solve 9sin^2theta - 6sintheta = 1 over [0 degree, 360 degrees) ?

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Answer 1 · 2018-03-31T11:44:26

#theta=53.58 color(white)(888)# and #color(white)(888)theta=126.42#

#theta=352.06 color(white)(888)# and # color(white)(888)theta=187.93 color(white)(888)# ( 2 d.p. )

#9sin^2(theta)-6sin(theta)=1#

#9sin^2(theta)-6sin(theta)-1=0#

This is just a quadratic equation, which we can easily solve:

Let # \ \ \ \u=sin(theta)#

Then:

#9u^2-6u-1=0#

Using quadratic formula:

#u=(-b+-sqrt(b^2-4ac))/(2a)#

#u=(-(-6)+-sqrt((-6)^2-(4(9)(-1))))/((2)(9))#

#u=(6+-sqrt(72))/(18)#

#u=(1+-sqrt(2))/3#

But # \ \ \ \ u=sin(theta)#

#sin(theta)=(1+sqrt(2))/3#

#sin(theta)=(1-sqrt(2))/3#

#theta=arcsin(sin(theta))=arcsin((1+sqrt(2))/3)#

#theta=arcsin(sin(theta))=arcsin((1-sqrt(2))/3)#

For:

#[0,360]#

#theta=arcsin((1+sqrt(2))/3)# and #theta=180-arcsin((1+sqrt(2))/3)#

#theta=360+arcsin((1-sqrt(2))/3)#

and

#theta=180-arcsin((1-sqrt(2))/3)#

Approximate values:

#theta=53.58 color(white)(888)# and #color(white)(888)theta=126.42#

#theta=352.06 color(white)(888)# and # color(white)(888)theta=187.93 color(white)(888)# ( 2 d.p. )