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EquationSolve the equation : sin^2x - 3sin2x + 5cos^2x = 0 .

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sodelete | Student, College Freshman | (Level 1) Honors

Posted June 3, 2011 at 12:05 AM via web

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Equation

Solve the equation :

sin^2x - 3sin2x + 5cos^2x = 0 .

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted June 3, 2011 at 12:36 AM (Answer #2)

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First, we'll recognize the type of the equation, that is a homogeneous equation in sin x and cos x. The method of solving is to divide entire equation by (cos x)^2.

Before dividing by (cos x)^2, we'll recall the double angle identity:

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

We'll 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 sreplace tan x by 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

All solutions of the equation are found in the following sets: {arctan 5 + k*pi}U{pi/4 + k*pi}.

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