Answer to Question #154815 in Trigonometry for .

Question

solve the following equation for 0<x<360

a) 6cos² x + sin x - 4

b)9 tan x + tan² x = 5 sec² x - 3

Answer 1

a) $6cos^2x+sinx-4=0$

let's use the following formula

$cos^2x+sin^2x=1$

Hence, $cos^2x=1-sin^2x$

Then, the equation can be written as follows:

$6(1-sin^2x)+sinx-4=0$

Let's assume that $sin x=u$ (hence $1\geq u\geq-1$ )

Then, the equation can be written as follows:

$6(1-u^2)+u-4=0$

$-6u^2+u+2=0$ (it is the quadratic equation).

Let's find its discriminant D:

$D=1^2-4*(-6)*2=1-(-48)=49=7^2>0$

Hence,

$u1=(-1+7)/(-6*2)=6/(-12)=-0.5$

$u2=(-1-7)/(-6*2)=-8/(-12)=2/3$

If $sin x = -0.5$ then $x1=210\degree$ and $x2=330\degree$

If $sinx=2/3$ then $x3=arcsin(2/3)=41.81\degree$ and $x4=arcsin(2/3)\degree+180=41.81\degree+180\degree=221.81\degree$

Answer: roots of the equation are following: $210\degree, 330\degree, 41.81\degree, 221.81\degree$

b)

$9tanx+tan^2x=5sec^2x-3$

$9*sinx/cosx+sin^2x/cos^2x=5/cos^2x-3$

Let's multyply both sides of an equation by $cos^2x$ :

$9sinx*cosx+sin^2x=5-3*cos^2x$

Let's use the following famous formulas:

$six2x=2sinxcosx$ and $cos2x=cos^2x-sin^2x=1-2sin^2x=2cos^2x-1$

Hence,

$9*sin2x/2=5-3cos^2x-(1-cos^2x)=4-2cos^2x=4-(2cos^2x-1)-1$

Then,

$4.5*sin2x=3-cos2x$ This equation can be rewritten as follows:

$(4.5*sin2x)^2=(3-cos2x)^2$

Hence,

$20.25sin^22x=9-6cos2x+cos^22x$

Then,

$20.25-20.25cos^22x=9-6cos2x+cos^22x$

$-21.25cos^22x+6cos2x+11,25=0$

Let's assume that $cos 2x=u$ (hence $-1\leq u \leq 1$ )

Then, the equation can be written as follows:

$-21.25u^2+6u+11.25=0$ (it is the quadratic equation).

Let's find its discriminant D:

$D=36-4*11.25*(-21.25)=992.25=31.5^2>0$

Hence,

$u1=(-6+31.5)/(2*(-21.25))=-0.6$

$u2=(-6-31.5)/(2*(-21.25))=0.8824$

if $cos2x=-0.6$ then $2x=arccos( -0.6)=323.14\degree$ or $216.86\degree$

hence $x = 161.57\degree$ or $x=108.43\degree$

if $cos2x=0.8824$ then $2x=arccos(0.8824)= 28.06\degree$ or $151.94\degree$

hence $x=14.03\degree$ or $x=75.97\degree$

The potential roots of the equation are following: $161.57\degree, 108.43\degree, 14.03\degree, 75.97\degree$.

In the solution both parts of the equation were squared. hence, It is necessary to check the occurence of extraneous roots.

Checking the root $161.57\degree$

$9(-0.333)+(-0.333)^2=5(-1.054)^2-3$ is not true, hence the root $161.57\degree$ is not correct.

Checking the root $108.43\degree$

$9(-3)+(-3)^2=5(-3.163)^2-3$ is not true, hence the root $161.57\degree$ is not correct.

Checking the root $14.03\degree$

$9(0.249)+(0.249)^2=5(1.03)^2-3$ is true, hence the root $161.57\degree$ is correct.

Checking the root $75.97\degree$

$9(4.001)+(4.001)^2=5(4.125)^2-3$ is not true, hence the root $161.57\degree$ is not correct.

Answer: the root of equation is $161.57\degree$