Science

Math

Humanities

... and beyond

Socratic Meta
Featured Answers

Topics

How do you solve $2 {cos}^{2} (3 x) + 5 cos (3 x) - 3 = 0$ $2cos^2(3x)+5cos(3x)-3=0$ ?

1 Answer

Aug 1, 2016

For $x \in [0, π]$ $x in [0,pi]$
$XXX x = \frac{π}{9}$ $color(white)("XXX")color(green)(x=pi/9)$

Explanation:

Let $θ = 3 x$ $theta=3x$
and $k = cos (θ)$ $k=cos(theta)$

The given equation becomes
$XXX 2 k^{2} + 5 k - 3 = 0$ $color(white)("XXX")2k^2+5k-3=0$

which can be factored as
$XXX (2 k - 1) (k + 3) = 0$ $color(white)("XXX")(2k-1)(k+3)=0$

with possible solutions
$XXX k = \frac{1}{2} XX or XX k = - 3$ $color(white)("XXX")k=1/2color(white)("XX")"or"color(white)("XX")k=-3$

but since $k = cos (θ)$ $k=cos(theta)$
$XXX k \in [- 1, + 1]$ $color(white)("XXX")k in [-1,+1]$
and $k = - 3$ $k=-3$ is an extraneous solution.

$k = cos (θ) = \frac{1}{2}$ $k=cos(theta)=1/2$ (for $θ \in [0, π]$ $theta in [0,pi]$ )
$\to θ = \frac{π}{3}$ $rarr theta =pi/3$

and since $3 x = θ$ $3x=theta$
$\to x = \frac{π}{9}$ $rarr x=pi/9$

Related questions

See all questions in Solving Trigonometric Equations

Impact of this question

4444 views around the world

You can reuse this answer
Creative Commons License