Question

Studying for Precalculus Test, and teaching myself is not working... The question states, "Find all solutions in the interval 0 degrees 10 years ago

Answer 1

\(\large 1 - 2sin\theta = 0\) on interval \(\large 0 \le \theta \le 2\pi\)

Focus first on the equation

\[\large 1 - 2sin\theta = 0\]

Subtract 1 from both sides and divide everything by -2
\[\large sin(\theta) = \frac{1}{2}\]

So now I will ask you...when does \(\large sin(\theta) = \frac{1}{2}\) ?

Answer 2

Oh! Would it be \[\pi/6, 5\pi/6, 7\pi/6, and 11\pi/6?\]

Answer 3

Yes and No..Remember we only want it within the boundaries of 0 and 2pi

Which of those are within this boundary :)

Answer 4

Uhm... well 2pi is the whole circle right? So... I'm confused lol

Answer 5

Lol okay lets try and do this out

Imagine going completely around the circle JUST ONCE!!! because going around once means you have gone 2pi

The 2 places where sin(theta) = 1/2 are at pi/6 and 5pi/6 as shown



There are no more parts where sin(theta) = 1/2 unless we start to go around the circle again

Answer 6

Ohhh okay! So when a question states interval of 0 to 2pi, its only the parts on half of the circle?

Answer 7

Not quite, but I didnt point that out

So our unit circle is just

Answer 8

Note here...we start where I labeled 2pi and move counter clockwise around the circle until we get back to that same point



That distance we just traveled going once fully around the circle...is 2pi

Answer 9

Ohh, and the other points would be negative, so you wouldnt include them... right?

Answer 10

So where they say the interval is 0 to 2pi....we start at that 2pi label....and go fully around the circle once back to that point...that is our interval

For example If the interval were instead 0 to pi/2 ....that would only be from



in that little area of the circle