If , then is equivalent to . Draw the unit circle and see for which angles cosine is negative.
Concerning , it helps to draw a graph of and see which portion is located between the lines y = 1 and y = 3. Can ever be greater than 1?
I was asked to solve from which the solution is or
The next part asks: Hence solve , where
I tried doing: Replace with
or
therefore, or but error
Can you show me the proper way to solve such an inequality? I totally forgot how to tackle this...
Yes, no value of can contribute to the solution of .
You are right. I missed that also implies , and even though the interval is outside of , the corresponding interval for : is inside .
It is better to be precise and to select the solutions from in the end. So, iff , . This is equivalent to , . Of these, only intervals for k = 0 and k = 1 lie inside .