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

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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

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Student Answers

giorgiana1976 | Student

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}.

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