Question

How do you solve \[\cos 2x = \dfrac{{\sqrt 2 }}{2}\] over the interval 0 to 2π?

Answer 1

Hint: This is a question of trigonometric solutions. Isolate the variable angle on the LHS and take all the operations to the RHS. The result you’ll get is the general solution. Now put the intervals in the general solution, to get the final answer.

Complete step by step solution:
The first step is to isolate the variable angle, “x” on the left hand side.
\[\cos 2x = \dfrac{{\sqrt 2 }}{2}\]
\[ \Rightarrow 2x = {\cos ^{ - 1}}(\dfrac{{\sqrt 2 }}{2})\]………….(Taking the cosine function from left to right hand side)
$ \Rightarrow x = \dfrac{1}{2}{\cos ^{ - 1}}(\dfrac{{\sqrt 2 }}{2})$……. Equation (i)
Now, ${\cos ^{ - 1}}(\dfrac{{\sqrt 2 }}{2}) = {\cos ^{ - 1}}(\dfrac{1}{{\sqrt 2 }})$.
We can also write, $\dfrac{1}{{\sqrt 2 }} = \cos ( \pm \dfrac{\pi }{4} + 2\pi n)$.
After putting the above value in equation (i),
$x = \dfrac{1}{2}{\cos ^{ - 1}}(\cos ( \pm \dfrac{\pi }{4} + 2\pi n)) \\
\Rightarrow x = \pm \dfrac{\pi }{8} + \pi n \\ $
Where n is any integer.
For \[n = 0\],
\[x = \dfrac{\pi }{8},\dfrac{{ - \pi }}{8}\]
Similarly, for \[n = 1,x = \dfrac{{9\pi }}{8},\dfrac{{7\pi }}{8}\] and \[n = 2,x = \dfrac{{17\pi }}{8},\dfrac{{15\pi }}{8}\]. Since the question asks for values in the interval of 0 to 2π, we need not calculate any further.

Therefore, the solutions for \[\cos 2x = \dfrac{{\sqrt 2 }}{2}\] in the interval of 0 to 2π are \[\dfrac{\pi }{8},\dfrac{{9\pi }}{8},\dfrac{{7\pi }}{8},\dfrac{{15\pi }}{8}\].

Note: There are 3 steps to follow when trying to find all solutions of any trigonometric equation. They are:
Step 1: Find the trigonometric values have to have to unravel the equation.
Step 2: Find all 'angles' that give us these values from step 1.
Step 3: Find the values of the unknown which will end in angles that we got here step 2.