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Algebra 2/ trig long answer help?

+1 vote
1)Find to the nearest degree the values of X in the interval 0degrees <= X < 360 degrees that satisfies 25cos^2 X -4=0

2) use the quadratic formula to find, to the nearest degree, all values of X in the interval 0degrees <= X < 360 degrees that satisfy 3cosX+1=1/cosX

3) find, to the nearest tenth of a radian, the possible values of <A in the interval 0<=A<2pi: 3sin^2A+3SinA-2=0
asked Feb 5, 2013 in TRIGONOMETRY by homeworkhelp Mentor

3 Answers

+2 votes

1). 25 cos2X - 4 = 0

Add 4 to each side.

25 cos2X = 4

Divide each side by 25.

cos2X = 4/25

It can be written as cos2X = (2/5)2

Apply square root each side.

√(cosX)2 = √(2/5)2

cos X = 2/5 = 0.4

Trigonometry table in cos66°.4' = 0.4003

cos X = cos66°.6' then X = 66°.4'.

The nearest degree the values of X is 66°.4'.

answered Feb 5, 2013 by richardson Scholar
+2 votes

2). 3cosX + 1 = 1/csoX

Multiply each side by 'cosX'.

cosX(3cosX + 1) = 1

3cos2X + cosX = 1

Subtract 1 from each side.

3cos2X + cosX -1 = 0.

Let cosX = t.

Then 3t2 + t - 1 = 0.

Comapre equation with standard from ax2 + bx + c = 0 and write the coefficients.

a = 3, b = 1, and c = -1.

The quadratic formula:t = [-b±√(b2-4ac)] / 2a.

Substitute a = 3, b = 1, and c = -1. in the quadratic formula.

t = [-1±√((1)2-4(3)(-1))] / 2(3)

t = [-1±√(1 + 12)] / 6 = [-1±√13] / 6.

Therefore t = (-1 + √13)/6 or t = ((-1 - √13)/6)

t = (-1+3.6055)/6 = 2.6055/6=0.4342

And t = (-1-3.6055)/6 = -4.6055/6 = -0.7675

But t = cosX so, cosX = 0.4342 or cosX = -0.7575

Trigonometry table in cos64°.3' = 0.4336 and cos 220°.9' = -0.7558

cosX = cos64°.3' then X = 64°.3'

And cosX = cos220°.9' then X = 220°.9'

The nearest degree the values of X is 66°.4' or 220°.9'

answered Feb 5, 2013 by richardson Scholar
+2 votes

3). 3sin2A + 3SinA - 2=0

Let sinA = x

3x2 + 3x - 2 = 0.

Compare equation with standard from ax2 + bx + c = 0 and write the coefficients.

a = 3, b = 3, and c = -2.

The quadratic formula x = [-b±√(b2-4ac)] / 2a.

Substitute a = 3, b = 3, and c = -2. in the quadratic formula.

x = [-3 ± √(3)2-4(3)(-2))] / 2(3).

x = [-3 ± √(9 + 24)] / 6 = [-3 ± √33] / 6.

x = [-3 + √33] / 6. or x = [-3 - √33] / 6.

x = [-3 + 5.7445]/6 = 2.7445/6 = 0.4574

And x = [-3 - 5.7445]/6 = -8.7445/6 = -1.4574

But x = sinA so, sinA = 0.4574 or sinA = -1.4574

The possible value is sinA = 0.4574

Trigonometry table in sin27°.2' = 0.4571.

Therefore sinA = sin27°.2' then A = 27°.2'.

The nearest tenth of a radian is 27°.2'.

answered Feb 5, 2013 by richardson Scholar

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