
How do you solve \[2{\cos ^2}x - \cos x - 1 = 0\] and find all the solutions in the interval \[\left[ {0,2\pi } \right)\]
Answer
498k+ views
Hint: We have to solve the given quadratic equation. To do that, we can easily solve it by using the formula of calculating the value of the variable by using the concept of determinants. We use the formula for solving the value of determinants. Then we just put in the value of determinant into the formula for finding the variable, and that gives us the answer. But it is a point to note that if the value of the determinant is less than 0, i.e., negative, then there is no possible solution for the given equation.
Formula Used:
We are going to use the formula of calculating the value of the variable by using Determinant:
Determinant, \[D = {b^2} - 4ac\]
and, \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
Complete step-by-step answer:
The given equation is \[2{\cos ^2}x - \cos x - 1 = 0\].
Let \[\cos x = k\].
Then, the given quadratic equation is \[2{k^2} - k - 1 = 0\].
First, we calculate the value of the determinant.
In this question, \[a = 2,{\rm{ }}b = - 1,{\rm{ }}c = - 1\].
The formula for finding determinant is:
\[D = {b^2} - 4ac\]
Putting in the values, we get,
$\Rightarrow$ \[D = {\left( { - 1} \right)^2} - 4 \times 2 \times \left( { - 1} \right) = 1 + 8 = 9 = {\left( 3 \right)^2}\]
Now, putting it in the formula to calculate the value of \[k\],
$\Rightarrow$ \[k = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( 3 \right)}^2}} }}{2} = \dfrac{{1 \pm 3}}{2} = - 1,2\]
Thus, \[\cos x = - \dfrac{1}{2}\] and \[\cos x = 1\]
Hence, solutions are \[\left\{ {0,\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}} \right\}\]
Note: In the given question, we needed to factorize the given polynomial. It is not true that whenever there is a given problem, there will be a solution. So, first we need to check if there are any solutions that can be possible. If there is, then only we proceed to calculate the solution. There is no point in calculating the solution, when there is not any. For instance, there is no solution to the point of intersection of two parallel lines. Two parallel lines never intersect, hence, can never have any point of intersection.
Formula Used:
We are going to use the formula of calculating the value of the variable by using Determinant:
Determinant, \[D = {b^2} - 4ac\]
and, \[x = \dfrac{{ - b \pm \sqrt D }}{{2a}}\]
Complete step-by-step answer:
The given equation is \[2{\cos ^2}x - \cos x - 1 = 0\].
Let \[\cos x = k\].
Then, the given quadratic equation is \[2{k^2} - k - 1 = 0\].
First, we calculate the value of the determinant.
In this question, \[a = 2,{\rm{ }}b = - 1,{\rm{ }}c = - 1\].
The formula for finding determinant is:
\[D = {b^2} - 4ac\]
Putting in the values, we get,
$\Rightarrow$ \[D = {\left( { - 1} \right)^2} - 4 \times 2 \times \left( { - 1} \right) = 1 + 8 = 9 = {\left( 3 \right)^2}\]
Now, putting it in the formula to calculate the value of \[k\],
$\Rightarrow$ \[k = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( 3 \right)}^2}} }}{2} = \dfrac{{1 \pm 3}}{2} = - 1,2\]
Thus, \[\cos x = - \dfrac{1}{2}\] and \[\cos x = 1\]
Hence, solutions are \[\left\{ {0,\dfrac{{2\pi }}{3},\dfrac{{4\pi }}{3}} \right\}\]
Note: In the given question, we needed to factorize the given polynomial. It is not true that whenever there is a given problem, there will be a solution. So, first we need to check if there are any solutions that can be possible. If there is, then only we proceed to calculate the solution. There is no point in calculating the solution, when there is not any. For instance, there is no solution to the point of intersection of two parallel lines. Two parallel lines never intersect, hence, can never have any point of intersection.
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