Solve giving values of theta from 0 to 360 degrees inclusive

1. 2sinx-cosecx=1

2. 2 tan(theta/2)+ 3 tan theta=0

Solve giving values of x from 0 to 2 pi inclusive:

1. tan2theta+ tantheta=0

2. 2cosec2theta=1 + tan^2

thanks if you can help:)

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- Oct 2nd 2006, 07:30 AMConfuzzled?Trig help solving.
Solve giving values of theta from 0 to 360 degrees inclusive

1. 2sinx-cosecx=1

2. 2 tan(theta/2)+ 3 tan theta=0

Solve giving values of x from 0 to 2 pi inclusive:

1. tan2theta+ tantheta=0

2. 2cosec2theta=1 + tan^2

thanks if you can help:) - Oct 2nd 2006, 08:13 AMSoroban
Hello, Confuzzled!

Here's the first one . . .

Quote:

Solve for 0° ≤ x ≤ 360°

1) .2·sin x - csc x .= .1

Multiply through by sin x: . 2·sin²x - 1 .= .sin x

We have: . 2·sin²x - sin x - 1 .= .0

which factors: . (sin x - 1)(2·sin x + 1) .= .0

and has these roots:

. . sin x - 1 .= .0 . →. sin x = 1 . → . x = 90°

. . 2·sinx + 1 .= .0 . → . sin x = -½ . → . x .= .210°, 330°

Quote:

2) .2·tan(x/2)+ 3 tan x .= .0

Using a double-angle identity, we have:

. . . . . . . . . . . . . .2·tan(x/2)

. . 2·tan(x/2) + 3 ---------------- .= .0

. . . . . . . . . . . . .1 - tan²(x/2)

This simplifies to: .2·tan³(x/2) - 8·tan(x/2) .= .0

which factors: . 2·tan(x/2) [tan²(x/2) - 4] .= .0

and has these roots:

. . 2·tan(x/2) = 0 . → . x/2 .= .0°, 180° . → . x .= .0°, 360°

. . tan²(x/2) = 4 . → . tan(x/2) = ±2 . → . x. = .2·arctan(±2) .≈ .126.9°, 233.1°