Question

How do you solve $\cos (2t) = \dfrac{1}{2}$ ?

Answer 1

Hint: Here, in this question, we are asked to solve $\cos (2t) = \dfrac{1}{2}$ . We will have to solve this problem by using the property of angles of cosine. The term $\cos (x) = \dfrac{1}{2}$ has two solutions, using that we will frame the expression and equation and then solve it.

Formula used: As per the properties of trigonometry,
For $\cos (x) = \dfrac{1}{2}$ , the two solutions are
\[\;x = \dfrac{\pi }{3} + 2k\pi \]
\[x = \dfrac{5}{3}\pi + 2k\pi \]

Complete step-by-step solution:
In this question, we are asked to solve $\cos (2t) = \dfrac{1}{2}$ . We will solve this problem by using the property of angles of cosine.
We know that $\cos (x) = \dfrac{1}{2}$ has two solutions, which are
\[\;x = \dfrac{\pi }{3} + 2k\pi \]
\[x = \dfrac{5}{3}\pi + 2k\pi \] ,
Given this as a solution, we need to simply substitute \[x \to 2t\] , and solve for $t$ .
So, substituting \[x \to 2t\] , we have
$\Rightarrow$\[\;2t = \dfrac{\pi }{3} + 2k\pi \]
$\Rightarrow$\[2t = \dfrac{5}{3}\pi + 2k\pi \]
From which we need to transfer the coefficient of $t$ to the other side by dividing the number $2$ to the other side, so we have
$\Rightarrow$\[\;t = \dfrac{1}{2}\left( {\dfrac{\pi }{3} + 2k\pi } \right)\]
$\Rightarrow$\[t = \dfrac{1}{2}\left( {\dfrac{5}{3}\pi + 2k\pi } \right)\]
And it becomes,
$\Rightarrow$\[\;t = \left( {\dfrac{\pi }{6} + k\pi } \right)\]
$\Rightarrow$\[t = \left( {\dfrac{5}{6}\pi + k\pi } \right)\]
We can still simplify the outcome by taking $\pi $ as common out,
$\Rightarrow$\[\;t = \pi \left( {\dfrac{1}{6} + k} \right)\]
$\Rightarrow$\[t = \pi \left( {\dfrac{5}{6} + k} \right)\]

Therefore, the solutions for $t$ are \[\;t = \pi \left( {\dfrac{1}{6} + k} \right)\]
\[t = \pi \left( {\dfrac{5}{6} + k} \right)\]

Note: The Law of Cosines relates to all three sides and one of the angles of an arbitrary (not necessarily right)
The property of angles of cosine:
\[\cos \left( {2n\pi {\text{ }} + {\text{ }}x} \right){\text{ }} = {\text{ }}\cos {\text{ }}x\] , where \[n \in Z\]
Further, it is observed that,
\[\cos {\text{ }}x = 0\] , when \[x{\text{ }} = {\text{ }} \pm \dfrac{\pi }{2},{\text{ }} \pm 3\dfrac{\pi }{2},{\text{ }} \pm 5\dfrac{\pi }{2},{\text{ }} \ldots \]
It means that \[\cos x\] vanishes when $x$ is an odd multiple of \[\dfrac{\pi }{2}\] .
So, \[\cos x = 0\] implies \[x = \left( {2n{\text{ }} + {\text{ }}1} \right)\dfrac{\pi }{2}\] , where $n$ takes the value of any integer.
In the same way, we can derive other values of \[\cos \] degrees like \[30^\circ ,{\text{ }}45^\circ ,{\text{ }}60^\circ ,{\text{ }}90^\circ ,{\text{ }}180^\circ ,{\text{ }}270^\circ \] and \[360^\circ \] .