Question 964778
Find an angle between 0 and 2pi that is coterminal with: (27pi)/10,
<pre>
I'll only do the first two, the other two are done the same way:


To find coterminal angles we add {{{2pi*n}}} to the angle, where
n is some integer, positive, negative, or zero.   

Therefore when we find n, the answer will be {{{(27pi)/10+2pi*n}}}

Since we want the coterminal angle to be between 0 and {{{2pi}}}, 
we write an inequality which indicates that:

{{{0<=(27pi)/10+2pi*n<2pi}}}

Divide through by {{{pi}}}

{{{0<=27/10+2*n<2}}}

Multiply through by 10

{{{0<=27+20n<20}}}

Subtract 27 from all three sides:

{{{-27<=20n<-7}}}

Divide through by 20

{{{-27/20<=n<-7/20}}}

{{{-1.35<=n<-0.35}}}

There is only one integer between -1.35 and -0.35, namely -1.

So n = -1, therefore the answer is

{{{(27pi)/10+2pi*n}}}
{{{(27pi)/10+2pi(-1)}}}
{{{(27pi)/10-2pi}}}
{{{(27pi)/10-2pi*expr(10/10)}}}
{{{(27pi)/10-20pi/10}}}
{{{(7pi)/10}}}

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</pre>coterminal with (-pi)/6,<pre> 

Same way:

We add {{{2pi*n}}} and the answer will be {{{(-pi)/6+2pi*n}}}, when we find n.

Since we want the coterminal angle to be between 0 and {{{2pi}}}, 
we write the inequality which indicates that:

{{{0<=(-pi)/6+2pi*n<2pi}}}

Divide through by {{{pi}}}

{{{0<=-1/6+2*n<2}}}

Multiply through by 6

{{{0<=-1+12n<12}}}

Add 1 to all three sides:

{{{1<=12n<13}}}

Divide through by 12

{{{1/12<=n<13/12}}}
      
{{{1/12<=n<1&1/12}}}

There is only one integer between {{{1/12}}} and {{{1&1/12}}}, namely 1.

So n = 1, therefore the answer is

{{{-pi/6+2pi*1}}}
{{{-pi/6+2pi}}}
{{{-pi/6+2pi*expr(6/6)}}}
{{{-pi/6+12pi/6}}}
{{{11pi/6}}}

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Edwin</pre>