Answer:
Option A is correct.
Solution for the given equation is, [tex]x = 0^{\circ}[/tex]
Step-by-step explanation:
Given that : [tex]2\cos^2x -\cos x -1 =0[/tex]
Let [tex]\cos x =y[/tex]
then our equation become;
[tex]2y^2-y-1= 0[/tex] .....[1]
A quadratic equation is of the form:
[tex]ax^2+bx+c =0[/tex].....[2] where a, b and c are coefficient and the solution is given by;
[tex]x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Comparing equation [1] and [2] we get;
a = 2 b = -1 and c =-1
then;
[tex]y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}[/tex]
Simplify:
[tex]y = \frac{ 1 \pm \sqrt{1+8}}{4}[/tex]
or
[tex]y = \frac{ 1 \pm \sqrt{9}}{4}[/tex]
[tex]y = \frac{ 1 \pm 3}{4}[/tex]
or
[tex]y = \frac{1+3}{4}[/tex] and [tex]y = \frac{1 -3}{4}[/tex]
Simplify:
y = 1 and [tex]y = -\frac{1}{2}[/tex]
Substitute y = cos x we have;
[tex]\cos x = 1[/tex]
⇒[tex]x = 0^{\circ}[/tex]
and
[tex]\cos x = -\frac{1}{2}[/tex]
⇒ [tex]x = 120^{\circ} \text{and} x = 240^{\circ}[/tex]
The solution set: [tex]\{0^{\circ}, 120^{\circ} , 240^{\circ}\}[/tex]
Therefore, the solution for the given equation [tex]2\cos^2x -\cos x -1 =0[/tex] is, [tex]0^{\circ}[/tex]
Answer:
The correct answer option is A. 0°.
Step-by-step explanation:
We are given a quadratic equation in cos which is factored in the same way like usual quadratic equation is factored:
[tex] 2cos^2x - cos x - 1 = 0 [/tex]
This can be factorized and can also be written as:
[tex] 2cos^2x - 2cosx + cosx - 1 = 0 [/tex]
or
[tex] 2cos x (cos x - 1) + 1(cos x - 1) = 0[/tex]
[tex] (2cos x + 1) (cos x - 1) = 0 [/tex]
[tex]2xcos+1=0[/tex] or [tex]cos x-1=0[/tex]
[tex]cos x = -\frac{1}{2}[/tex] or [tex]cos x = 1[/tex]
[tex]x = 120[/tex] or [tex]x = 0[/tex]
Therefore, the correct answer option is A. 0°.
