# Math Help - Finding solutions to an equation

1. ## Finding solutions to an equation

The question is the following:

Find all solutions of the equation cos x = 0 such that 0</x</2pi.
--my answer is: x=pi/2 and 3pi/2.

part b says:

Find all solutions of the equation cos x = 0.
--the solutions posted for this homework shows pi/2 + 2pi(k) where k is an integer and also 3pi/2 + 2pi(k). These two are then simplified to be x=pi/2 + pi(k).

BUT THEN...

for a similar question...

Find all solutions of the equation sin x = 1/2 such that 0</x</2pi
--my answer: pi/6 and 5pi/6

part b) find all solutions of the equation sin x = 1/2
--the homework answers have two answers: x=pi/6 + 2pi(k) and x=5pi/6 + 2pi(k)

and there is no simplification.

My question is how come there was simplification in the first question but not in the second one. And how did the first one get simplified to the final equation?

thank you, sorry for the long post!

2. ## Re: Finding solutions to an equation

Originally Posted by bhuang
The question is the following:

Find all solutions of the equation cos x = 0 such that 0</x</2pi.
--my answer is: x=pi/2 and 3pi/2.

part b says:

Find all solutions of the equation cos x = 0.
--the solutions posted for this homework shows pi/2 + 2pi(k) where k is an integer and also 3pi/2 + 2pi(k). These two are then simplified to be x=pi/2 + pi(k).

BUT THEN...

for a similar question...

Find all solutions of the equation sin x = 1/2 such that 0</x</2pi
--my answer: pi/6 and 5pi/6

part b) find all solutions of the equation sin x = 1/2
--the homework answers have two answers: x=pi/6 + 2pi(k) and x=5pi/6 + 2pi(k)

and there is no simplification.

My question is how come there was simplification in the first question but not in the second one. And how did the first one get simplified to the final equation?

thank you, sorry for the long post!
all solutions in the first equation differ by exactly $\pi$ radians ; the set of all solutions in the second example do not have such a "convenient" difference.