- #1

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Hello,

i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

Solve the following equations in the range -180°≤ θ ≤ 180°

tanθ + cotθ = 2

tanθ + cotθ = 2

[itex]\frac{sin\theta}{cos\theta}[/itex] + [itex]\frac{cos\theta}{sin\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta}{cos\theta sin\theta}[/itex] + [itex]\frac{cos^{2}\theta}{sin\theta cos\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

Since [itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

then

[itex]2cos\theta sin\theta[/itex] = 1

(the above is where I think i've done something careless)

I now have

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

I don't know what to do with it really... I tried saying [itex]sin^{2}\theta[/itex] = [itex] 1 - cos^{2}\theta[/itex] but it doesn't really get me anywhere.

Is what I have done not correct?

EDIT - I know it can be solved by putting the original equation into a quadratic form, and doing so did bring me to the correct answer of -135 and 45 degrees.

i'm fairly sure that what I have done is wrong/not allowed, but I want to check because it seemed like a good idea, but now it's causing problems.

## Homework Statement

Solve the following equations in the range -180°≤ θ ≤ 180°

tanθ + cotθ = 2

## Homework Equations

## The Attempt at a Solution

tanθ + cotθ = 2

[itex]\frac{sin\theta}{cos\theta}[/itex] + [itex]\frac{cos\theta}{sin\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta}{cos\theta sin\theta}[/itex] + [itex]\frac{cos^{2}\theta}{sin\theta cos\theta}[/itex] = 2

[itex]\frac{sin^{2}\theta + cos^{2}\theta}{cos\theta sin\theta}[/itex] = 2

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = [itex]2cos\theta sin\theta[/itex]

Since [itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

then

[itex]2cos\theta sin\theta[/itex] = 1

(the above is where I think i've done something careless)

I now have

[itex]sin^{2}\theta + cos^{2}\theta[/itex] = 1

I don't know what to do with it really... I tried saying [itex]sin^{2}\theta[/itex] = [itex] 1 - cos^{2}\theta[/itex] but it doesn't really get me anywhere.

Is what I have done not correct?

EDIT - I know it can be solved by putting the original equation into a quadratic form, and doing so did bring me to the correct answer of -135 and 45 degrees.

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