Question
Question: How do you solve \(\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2}\) over the interval \(0\) to \(2\...
How do you solve cos(t)=−23 over the interval 0 to 2π?
Solution
First, find the values of t satisfying cos(t)=−23 using trigonometric properties. Next, find all values of t in the interval [0,2π]. Then, we will get all solutions of the given equation in the given interval.
Formula used:
1. cos6π=23
2. cos(π−x)=−cosx
3. cos(π+x)=−cosx
Complete step by step solution:
Given equation: cos(t)=−23
We have to find all possible values of t satisfying a given equation over the interval 0 to 2π.
First, we will find the values of t satisfying cos(t)=−23…(i)
So, using the property cos(π−x)=−cosx and cos6π=23 in equation (i).
⇒cost=−cos6π
⇒cost=cos(π−6π)
⇒t=65π
Now, using the property cos(π+x)=−cosx and cos6π=23 in equation (i).
⇒cost=−cos6π
⇒cost=cos(π+6π)
⇒t=67π
Since, the period of the cost function is 2π so values will repeat every 2π radians in both directions.
t=65π+2nπ,67π+2nπ, for any integer n.
Now, find the values of n that produce a value within the interval [0,2π].
i.e., find all values of t over the interval 0 to 2π.
Plug in 0 for n and simplify to see if the solution is contained in [0,2π].
Since, it is given that t∈[0,2π], hence put n=0 in the general solution.
So, putting n=0 in t=65π+2nπ,67π+2nπ, we get
t=65π,67π
Thus, t=65π,67π or t=150∘,210∘.
Hence, t=65π,67π or t=150∘,210∘ are solutions of the given equation over the interval 0 to 2π.
Note: In above question, we can find the solutions of given equation by plotting the equation, cos(t)=−23 on graph paper and determine all solutions which lie in the interval, [0,2π].