solve the equation sin^2x-3sin2x+5cos^2x=0
2 Answers
hala718 | High School Teacher | (Level 1) Educator Emeritus
sin^2 x - 3sin2x + 5cos^2 x = 0
First let us rewrite:
We know that:
sin2x = 2sinx*cosx
Now substitute:
sin^2 x - 3*2sinx*cosx + 5cos^2 x = 0
sin^2 x - 6sinx*cosx + 5cos^2 x= 0
Nw let us factor:
(sin-5cosx)(sinx - cosx) = 0
Then we have two cases:
sinx -5cosx = 0
==> sinx= 5cosx
==> tanx = 5
==> x= arctan 5 + 2kpi.
Also,
sinx-cosx = 0
==> sinx = cosx
==> x= pi/4 + 2kpi
giorgiana1976 | College Teacher | (Level 3) Valedictorian
The given equation is a homogeneous equation in sin x and cos x and we'll divide the equation by (cos x)^2.
Before dividing by (cos x)^2, we'll write the formula for sin 2x:
sin 2x = sin (x+x) = sin x*cos x + sinx*cos x
sin 2x = 2 sin x*cos x
We'll substitute sin 2x by it's formula in the given equation:
(sin x)^2 - 3*2*sin x*cos x + 5 (cos x)^2 = 0
Now, we can divide by (cos x)^2:
(sin x/cos x)^2 - 6(sinx/cosx) + 5 = 0
But the ratio sin x/cos x = tan x
We'll substitute the ratio by the function tan x:
(tan x)^2 - 6tan x + 5 = 0
We'll substitute tan x = t
t^2 - 6t + 5 = 0
We'll apply the quadratic formula:
t1 = [6+sqrt(36 - 20)]/2
t1 = (6+4)/2
t1 = 5
t2 = 1
tan x = t1
tan x = 5
x = arctan 5 + k*pi
tan x = t2
tan x = 1
x = arctan 1 + k*pi
x = pi/4 + k*pi
The solution of the equation are: {arctan 5 + k*pi}U{pi/4 + k*pi}.