Question

How do you solve $\cos x = \dfrac{2}{3}\;\;\;;\csc x < 0$ ?

Answer 1

Now in order to solve the above equation we should work with one side at a time and manipulate it to the other side. Using some basic mathematical processes like we can simplify the above expression. In order to solve the given expression we have to use some trigonometric identities and mathematical processes we can solve the given equation.

Complete step by step solution:
Given
$\cos x = \dfrac{2}{3}{\text{ }};\csc x < 0............................\left( i \right)$
Here we have to basically solve for $x$ using any mathematical operations and functions. In order to solve for $x$ here we can see that the RHS is not given in any terms of $\pi$.
Such that we have used some other methods not involving the products of $\pi$ .
Now we have to solve for the variable $x$ such that the condition given is $\csc x < 0$.
So let’s take:

$$\cos x = \dfrac{2}{3} \\\ \Rightarrow x = {\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right).....................\left( {ii} \right) \\\$$

Now to solve (ii) we cannot find the value directly such that we can use calculator to find the value:
So we get:

$$\Rightarrow x = {\cos ^{ - 1}}\left( {\dfrac{2}{3}} \right) \\\ \Rightarrow x = {\cos ^{ - 1}}\left( {0.67} \right) \\\ \Rightarrow x = \pm {48.19^ \circ }\;\;.....................\left( {iii} \right) \\\$$

Now we have $x = \pm {48.19^ \circ }\;$ also with a condition $\csc x < 0$.
So since $\csc x < 0$ we can only accept the negative answers and thus have to neglect the positive one.
Therefore we can write on solving$\cos x = \dfrac{2}{3}\;\;\;;\csc x < 0$, we get $x = - {48.19^ \circ}$ which would be in Quadrant IV.

Note: While approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.