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How do you solve #Cos ( x + pi/6 ) = -0.5 # over the interval 0 to 2pi?

Answer 1

Daniel Acosta

#pi/2,(7pi)/6#

#cos(x+pi/6)=-1/2#

#=>cos(x+pi/6)=-cos(pi/3)#

#=>cos(x+pi/6)=cos(pi-pi/3)=cos(2/3pi)#

#=>x+pi/6=2/3pi#

#:.x=(2pi)/3-pi/6=pi/2#

Again

#=>cos(x+pi/6)=-cos(pi/3)#

#=>cos(x+pi/6)=cos(pi+pi/3)=cos(4/3pi)#

#=>x+pi/6=4/3pi#

#:.x=(4pi)/3-pi/6=(7pi)/6#

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Answer 2

Daniel Acosta

To solve ( \cos(x + \frac{\pi}{6}) = -0.5 ) over the interval ( 0 ) to ( 2\pi ):

Rewrite the equation as ( x + \frac{\pi}{6} = \arccos(-0.5) ).
Find the principal value of ( \arccos(-0.5) ) using a calculator or reference table. The principal value is ( \frac{7\pi}{3} ).
Subtract ( \frac{\pi}{6} ) from ( \frac{7\pi}{3} ) to find the solution for ( x ) within the given interval.
( x = \frac{7\pi}{3} - \frac{\pi}{6} = \frac{14\pi}{6} - \frac{\pi}{6} = \frac{13\pi}{6} ).
Check if ( \frac{13\pi}{6} ) lies within the interval ( 0 ) to ( 2\pi ).
Since ( \frac{13\pi}{6} ) is greater than ( 2\pi ), there are no solutions within the given interval.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.