How do you solve $3 {cos}^{2} x - 5 cos x - 1 = 3 {sin}^{2} x$ $3cos^2 x - 5cos x -1= 3 sin^2 x$ ?

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How do you solve $3 {cos}^{2} x - 5 cos x - 1 = 3 {sin}^{2} x$ $3cos^2 x - 5cos x -1= 3 sin^2 x$ ?

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Answer 1 · 2015-09-01T18:17:36

$x = 2 n π \pm 2 \frac{π}{3}$ $x = 2npi +- 2pi/3$

The description looks long, but I've written every single step so that anyone can understand!

Explanation:

$3 {cos}^{2} x - 5 cos x - 1 = 3 {sin}^{2} x$ $3cos^2x−5cosx−1=3sin^2x$

$3 ({cos}^{2} x - {sin}^{2} x) - 5 cos x - 1 = 0$ $3(cos^2x - sin^2x) - 5cosx-1=0$

$3 ({cos}^{2} x - 1 + {cos}^{2} x) - 5 cos x - 1 = 0$ $3(cos^2x-1+cos^2x) - 5cosx -1=0$

$6 {cos}^{2} x - 5 cos x - 4 = 0$ $6cos^2x -5cosx -4=0$

from here its just a quadratic equation. you divide it to its factors.

$(3 cos x - 4) (2 cos x + 1) = 0$ $(3cosx-4) (2cosx +1) =0$

$3 cos x - 4 = 0$ $3cosx-4=0$ or $2 cos x + 1 = 0$ $2cosx+1=0$
from here u get two possibilities

$cos x = \frac{4}{3}$ $cos x = 4/3$ or $cos x = - \frac{1}{2}$ $cos x = -1/2$
the first answer cannot be true because the cosines should be between -1 and +1
if u remember the cosine curve u should be able to understand what im talking about.

so we are left with the latter answer

$cos x = - \frac{π}{3}$ $cosx= -pi/3$
$cos x = (π - \frac{π}{3})$ $cos x = (pi - pi/3)$ you dont really have to show this step.. u can do it in your mind ;)
$cos x = 2 \frac{π}{3}$ $cos x =2pi/3$

according to the trig equations $θ = 2 n π \pm θ$ $theta = 2npi +- theta$
just replace the correct places of this equations with our answer
so the answer is
$x = 2 n π \pm 2 \frac{π}{3}$ $x = 2npi +- 2pi/3$

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How do you solve $3 {cos}^{2} x - 5 cos x - 1 = 3 {sin}^{2} x$ $3cos^2 x - 5cos x -1= 3 sin^2 x$ ?

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How do you solve 3cos^2 x - 5cos x -1= 3 sin^2 x3cos2x−5cosx−1=3sin2x3cos^2 x - 5cos x -1= 3 sin^2 x?

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How do you solve $3 {cos}^{2} x - 5 cos x - 1 = 3 {sin}^{2} x$ $3cos^2 x - 5cos x -1= 3 sin^2 x$ ?