Question

Which of the following is a solution to 2cos2x − cos x − 1 = 0?

Answer 1

Answer:

Option A is correct.

Solution for the given equation is, $x = 0^{\circ}$

Step-by-step explanation:

Given that : $2\cos^2x -\cos x -1 =0$

Let $\cos x =y$

then our equation become;

$2y^2-y-1= 0$           .....[1]

A quadratic equation is of the form:

$ax^2+bx+c =0$.....[2] where a, b and c are coefficient and the solution is given by;

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing equation [1] and [2] we get;

a = 2 b = -1 and c =-1

then;

$y = \frac{-(-1)\pm \sqrt{(-1)^2-4(2)(-1)}}{2(2)}$

Simplify:

$y = \frac{ 1 \pm \sqrt{1+8}}{4}$

or

$y = \frac{ 1 \pm \sqrt{9}}{4}$

$y = \frac{ 1 \pm 3}{4}$

or

$y = \frac{1+3}{4}$ and $y = \frac{1 -3}{4}$

Simplify:

y = 1 and $y = -\frac{1}{2}$

Substitute y = cos x we have;

$\cos x = 1$

⇒$x = 0^{\circ}$

and

$\cos x = -\frac{1}{2}$

⇒ $x = 120^{\circ} \text{and} x = 240^{\circ}$

The solution set:  $\{0^{\circ}, 120^{\circ} , 240^{\circ}\}$

Therefore, the solution for the given equation  $2\cos^2x -\cos x -1 =0$ is, $0^{\circ}$

Answer 2

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:

$2cos^2x - cos x - 1 = 0$

This can be factorized and can also be written as:

$2cos^2x - 2cosx + cosx - 1 = 0$

or

$2cos x (cos x - 1) + 1(cos x - 1) = 0$

$(2cos x + 1) (cos x - 1) = 0$

$2xcos+1=0$ or $cos x-1=0$

$cos x = -\frac{1}{2}$ or $cos x = 1$

$x = 120$ or $x = 0$

Therefore, the correct answer option is A. 0°.