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Solve each equation for X in the intervals 0</=X</=2pie?

+1 vote
1) 2Sin^2x-sinx=1 and #2) 25tan^x-70tanx=-49
asked Feb 21, 2013 in TRIGONOMETRY by andrew Scholar

2 Answers

+1 vote

1). 2Sin2(x) - sin(x) = 1.

Let sin(x) = t

2t2 - t = 1

Subtract 1 from each side.

2t2 - t - 1 = 0.

Now solve the equation using the factor method.

2t2 - 2t + t - 1 = 0.

2t(t - 1) +1(t - 1) = 0.

Take out common factors.

(2t + 1)(t - 1) = 0.

2t + 1 = 0 or t - 1 = 0.

2t + 1 = 0

Subtract 1 from each side. then 2t = -1.

Divide each side by 2. then t = -1/2

And t - 1 = 0.

Add 1 to each side.then t = 1.

Therefore t = -1/2 or t = 1

But t = sin(x)

sin(x)  = -1/2 or sin(x) = 1

Trigonometric table in sin(210°) = -1/2 and sin(90°) = 1

sin(x)  = sin(210°) or sin(x) = sin(90°)

x = 210° = 7π/6 or x = 90° = π/2.

Therefore x = π/6 + nπ or x = π/2 + nπ   [where n is an integer].

answered Feb 21, 2013 by britally Apprentice
+1 vote

2). 25tan(x) - 70tan(x) = -49.

Take out common term tan(x).

tan(x)[25 - 70] = - 49.

[-45] tan(x) = - 49

Multiply each side by negative one.

(45)tan(x) = 49

Divide each side by 45.

tan(x) = 49/45 = 1.088

Trigonometric table in tan(47°.42') = 1.088

tan(x) = tan(47°.42')

Apply tan-1 each side.

tan-1[tan(x)] = tan-1[tan(47°.42')]

Therefore x = 47°.42'.

answered Feb 21, 2013 by britally Apprentice

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