How do you solve ${cos}^{2} x - {sin}^{2} x = - cos x$ $cos^2x-sin^2x= -cosx$ ?

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How do you solve ${cos}^{2} x - {sin}^{2} x = - cos x$ $cos^2x-sin^2x= -cosx$ ?

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Answer 1 · 2016-03-03T09:45:44

It is

${cos}^{2} x - {sin}^{2} x = - cos x \Rightarrow {cos}^{2} x - (1 - {cos}^{2} x) = - cos x \Rightarrow 2 \cdot {cos}^{2} x + cos x - 1 = 0 \Rightarrow (cos x + 1) \cdot (2 \cdot cos x - 1) = 0$ $cos^2x-sin^2x= -cosx=>cos^2x-(1-cos^2x)=-cosx=> 2*cos^2x+cosx-1=0=>(cosx+1)*(2*cosx-1)=0$

Hence form the last equation we get

$cos x = - 1 \Rightarrow x = 2 \cdot k \cdot π \pm π$ $cosx=-1=>x=2*k*pi+-pi$ where $k$ $k$ is an integer

$2 \cdot cos x - 1 = 0 \Rightarrow cos x = \frac{1}{2} \Rightarrow x = 2 \cdot n \cdot π \pm \frac{π}{3}$ $2*cosx-1=0=>cosx=1/2=>x=2*n*pi+-pi/3$ where $n$ $n$ is an integer.

Finally the solutions are

$(2 \cdot k \cdot π \pm π, 2 \cdot n \cdot π \pm \frac{π}{3})$ $(2*k*pi+-pi,2*n*pi+-pi/3)$

Answer 2 · 2016-03-03T10:34:17

Final solution set

$x = {0^{\circ}, 60^{\circ}, 300^{\circ}, 180^{\circ}}$ $x = {0^@, 60^@,300^@, 180^@}$
Within the unit circle

Explanation:

Here is a simple
approach

we know ${cos}^{2} A - {sin}^{2} A = cos 2 A$ $cos^2 A - sin^2 A = cos 2A$

$- cos A = cos (- A)$ $- cosA = cos(-A)$

Using these we get;

${cos}^{2} x - {sin}^{2} x = - cos x$ $cos^2x-sin^2x= -cosx$

$cos 2 x = cos (- x)$ $cos 2x= cos (- x)$

$\Rightarrow 2 x = - x \Rightarrow 3 x = 0, x = 0$ $=> 2x = -x => 3x = 0 ,x = 0$

Right this is a definite solution

Lets go back to the equation

$2 {cos}^{2} x - 1 = - cos x$ $2cos^2 x - 1 = - cos x$

Bring everything over to one side

Let $cos x = a$ $cos x = a$

$2 a^{2} + a - 1 = 0$ $2a^2 + a -1 = 0$

Factoring you get

$(2 a - 1) (a + 1) = 0$ $(2a -1)(a + 1) = 0$

$2 a - 1 = 0$ $2a - 1 = 0$

$a = \frac{1}{2}$ $a = 1/2$

$\Rightarrow cos x = \frac{1}{2}$ $=>cos x = 1/2$

Now The first value which comes to mind is

$x = 60^{\circ}$ $x = 60^@$

Another value is

$x = 300^{\circ}$ $x = 300^@$

Lets repeat the same for the other part of the equation

$a + 1 = 0$ $a + 1 = 0$

$a = - 1$ $a = -1$

$\Rightarrow cos x = - 1$ $=>cos x =- 1$

There is only 1 possibility within a unit circle

$x = 180$ $x = 180$

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How do you solve ${cos}^{2} x - {sin}^{2} x = - cos x$ $cos^2x-sin^2x= -cosx$ ?

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Explanation:

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