Question

Solve the equation 5cos^2x-6tanx*cosx=cos^2x-5sin^2x-1.

Accepted Answer

We have to solve 5(cos x)^2 - 6 tan x*cos x = (cos x)^2- 5*( sin x)^2 - 1
5(cos x)^2 - 6 tan x*cos x = (cos x)^2- 5*( sin x)^2 - 1
=> 5(cos x)^2 + 5*( sin x)^2 - 6 tan x*cos x = (cos x)^2 - 1
we know (cos x)^2 + ( sin x)^2 = 1
=> 5 - 6 tan x*cos x = (cos x)^2 - 1
=> 6 - 6 (sin x/ cos x)* cos x =  (cos x)^2
=> 6 - 6 sin x = 1 - (sin x)^2
let y = sin x
=> 6 - 6y = 1 - y^2
=> y^2 - 6y+ 5 = 0
=> y^2 - 5y - y + 5 = 0
=> y( y - 5) - 1(y - 5) = 0
=> (y - 1)(y - 5) = 0
y = 1 or 5
As y = sin x, we can ignore x = 5.
So sin x = 1
x = arc sin (1)
x = pi/2 + 2*n*pi
Therefore x = pi/2 + 2*n*pi

Answer

For the beginning, we'll move all terms to one side:
5(cos x)^2 - 6tanx*cosx - (cos x)^2 + 5(sin x)^2 + 1 = 0
We'll re-group the terms:
[1 - (cos x)^2] - 6tanx*cosx + 5[(sin x)^2 + (cos x)^2] = 0
We'll write the fundamental formula of trigonometry;
[(sin x)^2 + (cos x)^2] = 1
(sin x)^2 = 1 - (cos x)^2
We also know that tan x = sin x/cos x
The equation will become:
(sin x)^2 - 6sinx*cos x/cosx + 5 = 0
We'll simplify and we'll get:
(sin x)^2 - 6sinx + 5 = 0
We'll substitute sin x = t
t^2 - 6t + 5 = 0
We'll apply quadratic formula:
t1 = [6 + sqrt(36 - 20)]/2
t1 = (6+4)/2
t1 = 5
t2 = (6-4)/2
t2 = 1
But sin x = t
sin x = t1
sin x = 5, impossible since sin x =< 1.
sin x = t2
sin x = 1
x = k*arcsin 1 + kpi
x = k*pi/2 + kpi
The solution of the equation is: {k*pi/2 + kpi}.

trigonometry1

Solve the equation 5cos^2x-6tanx*cosx=cos^2x-5sin^2x-1.

Expert Answers info

Unlock This Answer Now

Related Questions

Student Answers

Ask a Question

Popular Questions