Trigonometric Equation: Solve the equation (tanθ +1)(sin^2θ - sinθ) = 0 given that -π <= θ <= 2π

[tobyweir03]

Hello everyone,

My class for Maths Methods has just started for the year, and i'm struggling on a bit of homework in regards to trigonometric identities and equations. My teacher won't be able to reply to his emails this weekend for personal reasons, so I thought that i'd try asking everyone on the forums for some assistance. I've successfully completed all of this weekend's questions besides this one:

Solve the equation (tanθ +1)(sin^2θ - sinθ) = 0 given that -π <= θ <= 2π

I tried changing tanθ to sinθ/cosθ and what-not, but I can't seem to come up with an end product, sometimes leading to instances where I need to attempt to square root a negative number. I am certain I am doing something wrong, but I can't quite figure it out.

Any help would be greatly appreciated and taken into thoughtful consideration.

Thanks a million,
Toby

 

[MarkFL]

Factor as follows:

[MATH]\sin(\theta)(\tan(\theta)+1)(\sin(\theta)-1)=0[/MATH]
Now equate each factor to zero and in turn solve each over the given domain. What do you find?

 

[Deleted member 4993]

Hello everyone,

My class for Maths Methods has just started for the year, and i'm struggling on a bit of homework in regards to trigonometric identities and equations. My teacher won't be able to reply to his emails this weekend for personal reasons, so I thought that i'd try asking everyone on the forums for some assistance. I've successfully completed all of this weekend's questions besides this one:

Solve the equation (tanθ +1)(sin^2θ - sinθ) = 0 given that -π <= θ <= 2π

I tried changing tanθ to sinθ/cosθ and what-not, but I can't seem to come up with an end product, sometimes leading to instances where I need to attempt to square root a negative number. I am certain I am doing something wrong, but I can't quite figure it out.

Any help would be greatly appreciated and taken into thoughtful consideration.

Thanks a million,
Toby

First thing you should do is to realize that if:

(tanθ +1)(sin^2θ - sinθ) = 0

then

either...............................(tanθ +1) = 0

or/and.............................(sin^2θ - sinθ) = 0

Now you have two "simpler" equations to solve.

Continue........

 

[Harry_the_cat]

Welcome to the forum Toby. The phrase "maths methods' makes me think you are from Australia, as am I. Let us know whether you managed to finish off this problem, and good luck with your studies!

 

[Steven G]

Tanx + 1 = 0 => tanx =-1 => tanx =?
Another words do not do anything to tanx + 1

 

[Deleted member 4993]

First thing you should do is to realize that if:

(tanθ +1)(sin^2θ - sinθ) = 0

then

either...............................(tanθ +1) = 0

or/and.............................(sin^2θ - sinθ) = 0

Now you have two "simpler" equations to solve.

Continue........

Remember your domain of solution is -π <= θ <= 2π

Eache factor can result in multiple solutions.

 

[MarkFL]

Factor as follows:

[MATH]\sin(\theta)(\tan(\theta)+1)(\sin(\theta)-1)=0[/MATH]
Now equate each factor to zero and in turn solve each over the given domain. What do you find?

As 24 hours have passed without further feedback from the OP, I will follow up with my solution:

We must first observe that the tangent function is undefined for:

[MATH]\theta=\frac{\pi}{2}+k\pi=\frac{\pi}{2}(2k+1)[/MATH] where \(k\in\mathbb{Z}\)

[MATH]-\pi\le \frac{\pi}{2}(2k+1)\pi\le2\pi[/MATH]
[MATH]-2\le 2k+1\pi\le4[/MATH]
[MATH]-\frac{3}{2}\le k\pi\le\frac{3}{2}[/MATH]
Thus:

[MATH]k\in\{-1,0,1\}[/MATH]
And so we must discard any solution we find in the set:

[MATH]\theta\in\left\{-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2}\right\}[/MATH]
a) [MATH]\sin(\theta)=0[/MATH]
This implies:

[MATH]\theta=k\pi[/MATH]
Now, using the given domain:

[MATH]-\pi\le k\pi\le2\pi[/MATH]
[MATH]-1\le k\le2[/MATH]
Thus:

[MATH]k\in\{-1,0,1,2\}[/MATH]
And so:

[MATH]\theta\in\{-\pi,0,\pi,2\pi\}[/MATH]
b) [MATH]\tan(\theta)+1=0\implies \tan(\theta)=-1[/MATH]
This implies:

[MATH]\theta=\frac{3\pi}{4}+k\pi=\frac{\pi}{4}(4k+3)[/MATH]
[MATH]-\pi\le \frac{\pi}{4}(4k+3)\le2\pi[/MATH]
[MATH]-4\le 4k+3\le8[/MATH]
[MATH]-\frac{7}{4}\le k\le\frac{5}{4}[/MATH]
Thus:

[MATH]k\in\{-1,0,1\}[/MATH]
And so:

[MATH]\theta\in\left\{-\frac{\pi}{4},\frac{3\pi}{4},\frac{7\pi}{4}\right\}[/MATH]
c) [MATH]\sin(\theta)-1=0\implies \sin(\theta)=1[/MATH]
This implies:

[MATH]\theta=\frac{\pi}{2}+2k\pi=\frac{\pi}{2}(4k+1)[/MATH]
[MATH]-\pi\le \frac{\pi}{2}(4k+1)\le2\pi[/MATH]
[MATH]-2\le 4k+1\le4[/MATH]
[MATH]-\frac{3}{4}\le k\le\frac{3}{4}[/MATH]
Thus:

[MATH]k\in\{0\}[/MATH]
And so:

[MATH]\theta\in\left\{\frac{\pi}{2}\right\}[/MATH]
We must discard this solution.

And so the complete set of solutions over the given interval is:

[MATH]\theta\in\left\{-\pi,-\frac{\pi}{4},0,\frac{3\pi}{4},\pi,\frac{7\pi}{4},2\pi\right\}[/MATH]

 

[tobyweir03]

Thanks guys, I posted early because I didn't know what the speed of the forum post was. I checked in today and read all of your guides, including MarkFL's final explanation. I am indeed in Mathematics Methods from Western Australia, being in yr 11 means i have to juggle so I wasn't entirely focused on my Maths ?. I ended up with the same solution as you MarkFL, but I've yet to figure out why you discarded the value pi/2. Thanks everyone for your contributions, they are heavily appreciated.

EDIT: I presume the reason you withdrew pi/2 is because it fails to exist in negative form and doesn't follow a flow like the rest of the values? Mathematically I'm a little uncertain but that is just by following the patterns set.

 

[MarkFL]

The reason I discarded the value of [MATH]\theta=\frac{\pi}{2}[/MATH] is because the tangent function is undefined there, and causes the entire expression to become undefined. Essentially, it is not part of the domain of the expression, and so it cannot be considered a solution.