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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)=32\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2} over the interval 00 to 2π2\pi?

Explanation

Solution

First, find the values of tt satisfying cos(t)=32\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2} using trigonometric properties. Next, find all values of tt in the interval [0,2π]\left[ {0,2\pi } \right]. Then, we will get all solutions of the given equation in the given interval.
Formula used:
1. cosπ6=32\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2}
2. cos(πx)=cosx\cos \left( {\pi - x} \right) = - \cos x
3. cos(π+x)=cosx\cos \left( {\pi + x} \right) = - \cos x

Complete step by step solution:
Given equation: cos(t)=32\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2}
We have to find all possible values of tt satisfying a given equation over the interval 00 to 2π2\pi.
First, we will find the values of tt satisfying cos(t)=32\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2}…(i)
So, using the property cos(πx)=cosx\cos \left( {\pi - x} \right) = - \cos x and cosπ6=32\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2} in equation (i).
cost=cosπ6\Rightarrow \cos t = - \cos \dfrac{\pi }{6}
cost=cos(ππ6)\Rightarrow \cos t = \cos \left( {\pi - \dfrac{\pi }{6}} \right)
t=5π6\Rightarrow t = \dfrac{{5\pi }}{6}
Now, using the property cos(π+x)=cosx\cos \left( {\pi + x} \right) = - \cos x and cosπ6=32\cos \dfrac{\pi }{6} = \dfrac{{\sqrt 3 }}{2} in equation (i).
cost=cosπ6\Rightarrow \cos t = - \cos \dfrac{\pi }{6}
cost=cos(π+π6)\Rightarrow \cos t = \cos \left( {\pi + \dfrac{\pi }{6}} \right)
t=7π6\Rightarrow t = \dfrac{{7\pi }}{6}
Since, the period of the cost\cos t function is 2π2\pi so values will repeat every 2π2\pi radians in both directions.
t=5π6+2nπ,7π6+2nπt = \dfrac{{5\pi }}{6} + 2n\pi ,\dfrac{{7\pi }}{6} + 2n\pi , for any integer nn.
Now, find the values of nn that produce a value within the interval [0,2π]\left[ {0,2\pi } \right].
i.e., find all values of tt over the interval 00 to 2π2\pi.
Plug in 00 for nn and simplify to see if the solution is contained in [0,2π]\left[ {0,2\pi } \right].
Since, it is given that t[0,2π]t \in \left[ {0,2\pi } \right], hence put n=0n = 0 in the general solution.
So, putting n=0n = 0 in t=5π6+2nπ,7π6+2nπt = \dfrac{{5\pi }}{6} + 2n\pi ,\dfrac{{7\pi }}{6} + 2n\pi , we get
t=5π6,7π6t = \dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6}
Thus, t=5π6,7π6t = \dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6} or t=150,210t = {150^ \circ },{210^ \circ }.

Hence, t=5π6,7π6t = \dfrac{{5\pi }}{6},\dfrac{{7\pi }}{6} or t=150,210t = {150^ \circ },{210^ \circ } are solutions of the given equation over the interval 00 to 2π.2\pi.

Note: In above question, we can find the solutions of given equation by plotting the equation, cos(t)=32\cos \left( t \right) = - \dfrac{{\sqrt 3 }}{2} on graph paper and determine all solutions which lie in the interval, [0,2π]\left[ {0,2\pi } \right].