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Solve the equation: 2 sin 2x + 3*(sin x+cos x) = -2.

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joemauv | Student, College Freshman | eNoter

Posted January 31, 2011 at 4:16 PM via web

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Solve the equation: 2 sin 2x + 3*(sin x+cos x) = -2.

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giorgiana1976 | College Teacher | Valedictorian

Posted January 31, 2011 at 4:28 PM (Answer #1)

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We'll solve this equation algebraically.

We'll move all terms to one side:

2sin2x + 3(sinx+cosx) + 2 = 0

We'll substitute sin 2x = 2sin x*cos x

2*2sin x*cos x + 3(sinx+cosx) + 2 = 0

We'll note sin x + cos x = y.

We'll raise to square and we'll get:

(sin x + cos x)^2 = y^2

(sin x)^2 + (cos x)^2 + 2sin x*cos x = y^2

But (sin x)^2 + (cos x)^2 = 1:

1 + 2sin x*cos x = y^2

2sin x*cos x = y^2 - 1

We'll re-write the given equation in y:

2*(y^2 - 1) + 3(y) + 2 = 0

We'll remove the brackets:

2y^2 - 2 + 3y + 2 = 0

We'll eliminate like terms:

2y^2 + 3y = 0

We'll factorize by y:

y(2y + 3) = 0

We'll put y = 0

But y = sin x + cos x:

sin x + cos x = 0

We'll divide by cos x:

tan x + 1 = 0

tan x = -1

x = -pi/4 + kpi

2y + 3 = 0

2y = -3

y = -3/2

But the range of values of y is [-2;2].

Maximum of the sum: sin x + cos x = 1 + 1 = 2

Minimumof the sum: sin x + cos x = -1-1 = -2

We'll work with substitution:

sin x = 2t/(1+t^2)

cos t = (1-t^2)/(1+t^2)

2t/(1+t^2) + (1-t^2)/(1+t^2) = -3/2

4t + 2 - 2t^2 = -3 - 3t^2

We'll move all terms to one side:

t^2 + 4t + 5 = 0

t1 = [-4+sqrt(16-20)]/2

Since delta is negative, the equation has no real solutions.

The only solution of the equation is: x = -pi/4 + kpi.

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