find all real numbers in this interval (o, 2 pi) that satisfy this equation. round to the nearest tenth. cos x(square root symbol) 3 = 3(square root symbol)3 sin xi've been working on this all night!!

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If the equation is cos x*sqrt 3 = 3*sqrt (3*sin x), we'll raise to square both sides to remove the square root from both sides:

3(cos x)^2 = 9*3*sin x

We'll divide by 3 both sides:

(cos x)^2 = 9*sin x

We'll use the Pythagorean identity to express cos x with respect to sin x:

(cos x)^2 = 1 - (sin x)^2

1 - (sin x)^2 = 9*sin x

We'll subtract 9sin x both sides:

- (sin x)^2 - 9*sin x + 1 = 0

(sin x)^2 + 9*sin x - 1 = 0

We'll replace sin x by t:

t^2 + 9t - 1 = 0

We'll apply quadratic formula:

t1 = [-9 + sqrt(9^2 - 4*1*(-1))]/2*1

t1 = (-9+sqrt85)/2

t1 = 0.1097

t2 = (-9-sqrt85)/2

t2 = -9.1097

sin x = t1 => sin x = 0.1097 => x = arcsin 0.1097 => x = 6.2980 degrees (1 st quadrant) or x = 180 - 6.2980 = 173.702 degrees (2 nd quadrant)

sin x = -9.1097 impossible since the values of the function sine cannot be smaller than -1.

Therefore, the solutions of the equation are expressed in degrees and they are: {6.2980 ; 173.702}.

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