Please help me solve the equation?

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Answer 1 · 2018-05-11T20:57:43

See below.

1).

#4cos^2(theta)=1#

#cos^2(theta)=1/4#

Taking roots:

#cos(theta)=+-sqrt(1/4)=1/2#

#arccos(cos(theta))=arccos(1/2)=>theta=60^@,300^@#

#arccos(cos(theta))=arccos(-1/2)=>theta=120^@,240^@#

Solutions:

#60^@,120^@,240^@,300^@#

2).

#cos^2(theta)+cos(theta)-1=0#

Let #u=cos(theta)#

Then:

#u^2+u-1=0#

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

#u=-1/2+-sqrt(5)/2#

#cos(theta)=-1/2+sqrt(5)/2#

#arccos(cos(theta))=arccos(-1/2+sqrt(5)/2)#

This is in quadrant I, we need go no further because of the interval given.

#cos(theta)=-1/2-sqrt(5)/2#

#arccos(cos(theta))=arccos(-1/2-sqrt(5)/2)#

This is undefined for real numbers.

#-1<=cos(theta)<=1#

So:

#0^@<=arccos<=180^@#

#-1/2-sqrt(5)/2~~-1.618033988#

So solution is just:

#arccos(-1/2+sqrt(5)/2)~~51.8297621^@#