Question

How do you solve \[2{\cos ^2}x - \cos x = 1\] and find all solutions in the interval \[0 \leqslant x < 360\]?

Answer 1

Hint: Here the given question is in the form of trigonometric equations. First solve the given trigonometric equation by factorization method and next to find the angles by equating the trigonometric values to the RHS i.e., 0 and on further simplification we get the values of the unknown that will result in angles.

Complete step-by-step answer:
Factorization means the process of creating a list of factors otherwise in mathematics, factorization or factoring is the decomposition of an object into a product of other objects, or factors, which when multiplied together give the original.

The general form of an equation is \[a{x^2} + bx + c\] when the coefficient of \[{x^2}\] is unity. Every quadratic equation can be expressed as \[{x^2} + bx + c = \left( {x + d} \right)\left( {x + e} \right)\]. Here, b is the sum of d and e and c is the product of d and e.

Consider the given trigonometric equation
\[ \Rightarrow \,\,\,2{\cos ^2}x - \cos x = 1\]
Take RHS value 1 to the LHS, then the equation can be written as
 \[ \Rightarrow \,\,\,2{\cos ^2}x - \cos x - 1 = 0\]
Applying the factorization method Break the middle term as the summation of two numbers such that its product is equal to \[ - 2{\cos ^2}x\]. Calculated above such two numbers are \[ - 2\cos x\] and \[\cos x\].
\[ \Rightarrow \,\,\,2{\cos ^2}x - 2\cos x + \cos x - 1 = 0\]
Making pairs of terms in the above expression
\[ \Rightarrow \,\,\,\left( {2{{\cos }^2}x - 2\cos x} \right) + \left( {\cos x - 1} \right) = 0\]
Take out greatest common divisor GCD from the both pairs, then
\[ \Rightarrow \,\,\,2\cos x\left( {\cos x - 1} \right) + 1\left( {\cos x - 1} \right) = 0\]
Take \[\left( {\cos x - 1} \right)\] common
\[ \Rightarrow \,\,\,\left( {\cos x - 1} \right)\left( {2\cos x + 1} \right) = 0\]
Equate each term to the RHS 0
\[ \Rightarrow \,\,\,\left( {\cos x - 1} \right) = 0\] or \[\left( {2\cos x + 1} \right)\]
\[ \Rightarrow \,\,\,\cos x = 1\] or \[2\cos x = - 1\]
\[ \Rightarrow \,\,\,\cos x = 1\] or \[\cos x = - \dfrac{1}{2}\]
\[ \Rightarrow \,\,\,x = {\cos ^{ - 1}}\left( 1 \right)\]
\[\therefore \,\,\,\,x = 0\,\,\,\forall \,\,0 \leqslant x < 360.\]
and
\[ \Rightarrow \,\,\,x = {\cos ^{ - 1}}\left( { - \dfrac{1}{2}} \right)\]
\[\therefore \,\,\,\,x = 180 \pm 60\, = 120\,,\,240\]
\[\therefore \,\,\,\,x = 120,240\,\,\,\forall \,\,0 \leqslant x < 360.\]

Hence, The solution set of the equation \[2{\cos ^2}x - \cos x = 1\] is \[\left\{ {0,120,240} \right\} \subset \left[ {0,360} \right)\].

Note: The given function in this question is of the form of trigonometry function. When we see the given function it is in the form quadratic equation. We solve the quadratic equation by factorisation. To determine the exact value of x we use the table of trigonometry ratios for the standard angles.