How do you solve $11 = 3 {csc}^{2} x + 7$ $11=3csc^2x+7$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2016-09-04T07:27:54

$x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$ $x=pi/3,(2pi)/(3),(4pi)/(3),(5pi)/(3)$

We have: $11 = 3 {csc}^{2} (x) + 7$ $11=3csc^(2)(x)+7$ ; $0 \leq x \leq 2 π$ $0leqxleq2pi$

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

$\Rightarrow 3 {csc}^{2} (x) = 4$ $=> 3csc^(2)(x)=4$

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

$\Rightarrow {csc}^{2} (x) = \frac{4}{3}$ $=> csc^(2)(x)=4/3$

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

$\Rightarrow csc (x) = \pm \sqrt{\frac{4}{3}}$ $=> csc(x)=pmsqrt(4/3)$

$\Rightarrow csc (x) = \pm \frac{2 \sqrt{3}}{3}$ $=> csc(x)=pm(2sqrt(3))/3$

$\Rightarrow x = \frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}, \frac{5 π}{3}$ $=> x=pi/3,(2pi)/(3),(4pi)/(3),(5pi)/(3)$

How do you solve 11=3csc^2x+711=3csc2x+711=3csc^2x+7 for 0<=x<=2pi0≤x≤2π0<=x<=2pi?