Question

How do you solve $\sin \left( {x + {{60}^ \circ }} \right) + \cos \left( {x + {{30}^ \circ }} \right) = \dfrac{1}{2}$ in the range ${0^ \circ } < x < {360^ \circ }$ ?

Answer 1

Hint: The question belongs to the trigonometric equation. In this type of problem we will use the sum or difference formula of trigonometric identities. We will use a standard table of trigonometric ratios values at different angles. We know that both the trigonometric ratios sin and cos are periodic functions with periods equal to $2\pi $ . The domain of $\sin $ function is defined in the interval of $\left( { - \infty ,\infty } \right)$and the range of$\sin $function is $\left[ { - 1,1} \right]$ . $\operatorname{Sin} $ Is an odd function which means its graph will be symmetric about the origin. An odd function is defined as a function which is symmetric about the origin. The maximum points of the function are $\left( {\dfrac{\pi }{2} + 2n\pi ,1} \right)$ where $n$ is an integer and the minimum points of the function are $\left( {\dfrac{{3\pi }}{2} + 2n\pi , - 1} \right)$ where $k$ is an integer.

Complete step by step solution:
Step: 1 the giver trigonometric equation is,
$\sin \left( {x + {{60}^ \circ }} \right) + \cos \left( {x + {{30}^ \circ }} \right) = \dfrac{1}{2}$
Now use the sum formula of the trigonometric equation to solve the given equation,
We know that the sum formula of the trigonometric ratios are defined as,
$
  \sin \left( {a + b} \right) = \sin a\cos b - \cos a\sin b \\
  \cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b \\
 $
 Step: 2 use the formula in the given trigonometric equation.
$
  \sin \left( {x + {{60}^ \circ }} \right) + \cos \left( {x + {{30}^ \circ }} \right) = \dfrac{1}{2} \\
   \Rightarrow \sin x\cos 60 + \sin 60\cos x + \cos x\cos 30 - \sin x\sin 30 = \dfrac{1}{2} \\
 $
Substitute the values of standard trigonometric ratio in the equation.
$
   \Rightarrow \sin x\cos 60 + \sin 60\cos x + \cos x\cos 30 - \sin x\sin 30 = \dfrac{1}{2} \\
   \Rightarrow \sin x \times \dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}\cos x + \cos x \times \dfrac{{\sqrt 3 }}{2} - \sin x \times \dfrac{1}{2} = \dfrac{1}{2} \\
 $
Step: 3 add the similar term in the equation and cancel out the same term, having opposite signs.
$ \Rightarrow \sqrt 3 \cos x = \dfrac{1}{2}$
Step: 4 solve the equation to find the value of $x$ .
$ \Rightarrow \cos x = \dfrac{1}{{2\sqrt 3 }}$
Multiply the $\sqrt 3 $ in both denominator and numerator of the fraction.
$ \Rightarrow \cos x = \dfrac{{\sqrt 3 }}{{2\sqrt 3 \times \sqrt 3 }}$
Simplify the fraction to find the value of $x$ .
$
   \Rightarrow \cos x = \dfrac{{\sqrt 3 }}{6} \\
   \Rightarrow x = {73.22^ \circ } \\
 $
The value of the equation in the given range is equal to,
$
   \Rightarrow x = 360 - 73.22 \\
   \Rightarrow x = 286.78 \\
 $

Therefore the value of the equation is 286.78.

Note:
Use the sum formula of trigonometric ratio to solve the given equation. Students are advised to remember the standard table of trigonometric ratios. After getting the value of $x$ , subtract it from the 360. They must know to solve the trigonometric equation.