Question

How do you solve: $ \sin x = - 0.78 $ on $ 0 \leqslant x \leqslant 2\pi $ ?

Answer 1

Hint: In order to determine the values of $ x $ in the given equation, first use the calculator to find the value of $ x $ which gives the value $ 0.78 $ . Then look out for the range. Since, the value of $ \sin x $ is negative that means it lies in the 3rd and 4th Quadrant as $ \sin x $ is always negative in these Quadrants.

Complete step by step solution:
We are given the equation $ \sin x = - 0.78 $ .
Since, we know that there is no recognisable value of sine at any angle which gives $ 0.78 $ as a result. So, we need to use a calculator and according to that $ \sin {51.78^ \circ } = 0.78 $ .
Therefore, the value of $ x = {51.78^ \circ } $ .
But according to the question given the sign of the value is negative and for sine, we know that it can be negative in the 3rd of 4th Quadrants.
Let’s solve for value of sign in the 3rd quadrant and we get:
 $
  \sin x = - 0.78 \\
  \sin \left( {180 + x} \right) = - 0.78 \\
  \sin {(180 + 51.78)^ \circ } = - 0.78 \\
  \sin {231.78^ \circ } = - 0.78 \;
  $
So, for the 3rd Quadrant the value of sine is: $ {231.78^ \circ} $
Now, let's check for 4th Quadrant and we get:
 $
  \sin x = - 0.78 \\
  \sin \left( {360 - x} \right) = - 0.78 \\
  \sin {(360 - 51.78)^ \circ } = - 0.78 \\
  \sin {308.22^ \circ } = - 0.78 \;
  $
So, for the 4th Quadrant the value of sine is: $ {308.22^ \circ } $
Hence, we conclude that for the range $ 0 \leqslant x \leqslant 2\pi $ the value of sine which gives $ \sin x = - 0.78 $ as results are $ {231.78^ \circ } $ for the 3rd Quadrant and $ {308.22^ \circ } $ for the 4th Quadrant.

Note: Always look for the range before solving the equation. In which range the values will be negative or positive.
If there is no recognisable angle and their respective trigonometric value, then use of a calculator is needed for ease.
For other trigonometric values same methods would be followed but their Quadrants would be different.