Solve the equation 3 csc x − 2 = 0 \sqrt{3\csc x} - 2 = 0 3 csc x − 2 = 0 .
Solution.
3 csc ( x ) − 2 = 0 ⇒ 3 csc ( x ) = 2 ⇒ 3 csc ( x ) = 4 ⇒ csc ( x ) = 4 3 ⇒ sin x = 3 4 ⇒ ⇒ [ x = arcsin ( 3 4 ) + 2 π k , k ∈ Z x = π − arcsin ( 3 4 ) + 2 π k , k ∈ Z \begin{aligned}
\sqrt{3\csc(x)} - 2 &= 0 \Rightarrow \sqrt{3\csc(x)} = 2 \Rightarrow 3\csc(x) = 4 \Rightarrow \csc(x) = \frac{4}{3} \Rightarrow \sin x = \frac{3}{4} \Rightarrow \\
&\Rightarrow \left[ \begin{array}{l} x = \arcsin\left(\frac{3}{4}\right) + 2\pi k, \, k \in \mathbb{Z} \\ x = \pi - \arcsin\left(\frac{3}{4}\right) + 2\pi k, \, k \in \mathbb{Z} \end{array} \right.
\end{aligned} 3 csc ( x ) − 2 = 0 ⇒ 3 csc ( x ) = 2 ⇒ 3 csc ( x ) = 4 ⇒ csc ( x ) = 3 4 ⇒ sin x = 4 3 ⇒ ⇒ [ x = arcsin ( 4 3 ) + 2 πk , k ∈ Z x = π − arcsin ( 4 3 ) + 2 πk , k ∈ Z
Answer.
U k ∈ Z { arcsin ( 3 4 ) + 2 π k , π − arcsin ( 3 4 ) + 2 π k } . U_{k \in \mathbb{Z}} \left\{ \arcsin\left(\frac{3}{4}\right) + 2\pi k, \, \pi - \arcsin\left(\frac{3}{4}\right) + 2\pi k \right\}. U k ∈ Z { arcsin ( 4 3 ) + 2 πk , π − arcsin ( 4 3 ) + 2 πk } .