How do you solve #11=3csc^2x+7# for #0<=x<=2pi#?

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Answer 1 · 2016-09-04T07:27:54

#x=pi/3,(2pi)/(3),(4pi)/(3),(5pi)/(3)#

We have: #11=3csc^(2)(x)+7#; #0leqxleq2pi#

Let's subtract #7# from both sides of the equation:

#=> 3csc^(2)(x)=4#

Then, we can divide both sides by #3#:

#=> csc^(2)(x)=4/3#

To simplify the left-hand side we can take the square root of both sides:

#=> csc(x)=pmsqrt(4/3)#

#=> csc(x)=pm(2sqrt(3))/3#

#=> x=pi/3,(2pi)/(3),(4pi)/(3),(5pi)/(3)#