Question

Solve the given expression of trigonometric function $\cos \left( {\dfrac{\theta }{3} - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}$?

Answer 1

Here we have to solve for the variable theta, for this we have to solve the right side of equation first and make it to “cos” term, the given numerical value can be converted as in “cos” function, then on simple comparison you can get the value for the variable theta. For conversion into “cos” you should know the value of the “cos” function for the numerical value.

Formulae Used:
$\dfrac{1}{2} = \cos \left( {\dfrac{\pi }{3}} \right)$

Complete step by step solution:
The given question is $\cos \left( {\dfrac{\theta }{3} - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}$
To solve this question first we need to solve for the right side of the equation and get the value of “cos” function for the numerical value,
On solving we get:
$\Rightarrow \dfrac{1}{2} = \cos \left( {\dfrac{\pi }{3}} \right)$
Now putting the value obtained above in our main equation we get:
$\Rightarrow \cos \left( {\dfrac{\theta }{3} - \dfrac{\pi }{3}} \right) = \cos \left( {\dfrac{\pi }{3}}\right)$
On solving by removing “cos” function from both side of equation we get:

$$\Rightarrow \left( {\dfrac{\theta }{3} - \dfrac{\pi }{3}} \right) = \left( {\dfrac{\pi }{3}} \right) \\\ \Rightarrow \dfrac{\theta }{3} = \dfrac{\pi }{3} + \dfrac{\pi }{3} = \dfrac{{2\pi }}{3} \\\ \Rightarrow \theta = \dfrac{{2\pi }}{3} \times 3 = 2\pi \\\$$

Here we get the solution for the variable theta.

Additional Information: The given question can be solved by the comparison because there is one unknown variable theta, for which you have to compare the both sides of the equation needed to be compared only.

Note: Here we can also use the approach of differentiation on both side which will remove the “cos” function and bring “sin” function and on the other side of the equation the numerical value will become zero then simply transferring the “sin” function on the right side of the equation we can get the value of “sin0” and then solve further, but this approach will also take the same number of steps what we are using here.