# Solve acos²θ+bsinθ+c=0 for all values 0≤θ≤360°

## Homework Statement

Solve acos²θ+bsinθ+c=0 for all values 0≤θ≤360°

a=16
b=6
c=-12

So 16cos²θ+6sinθ-12=0

Cos²x=1-Sin²x

## The Attempt at a Solution

Identity: Cos²x=1-Sin²x

16(1-Sin²θ)+6Sinθ-12=0

16-16Sin²θ+6Sinθ-12=0

6Sinθ-16Sin²θ=12-16=-4

Divide by 2(?)

3Sinθ-8Sin²θ=-2
3Sinθ-4Sin³θ=-2

3Sinθ-4Sin³θ=Sin3θ

Sin3θ=-2

I realised that I am stuck at this point and can’t work out how to find θ from this, although it feels like I am missing something obvious. Any help would be appreciated!

Related Precalculus Mathematics Homework Help News on Phys.org
3Sinθ-8Sin²θ=-2
3Sinθ-4Sin³θ=-2
$$3\sinθ-8\sin^2θ=-2$$
$$3\sinθ-4\sin^{\color{red}{3}}θ=-2$$

Where does that three come from ?

$$3\sinθ-8\sin^2θ=-2$$
$$3\sinθ-4\sin^{\color{red}{3}}θ=-2$$

Where does that three come from ?
I tried to change 8Sin²θ into Sin³θ, but I guess that doesn’t work. I think it’s because I had Sin3θ written down, so I was trying to make it work out as that.

16-16Sin²θ+6Sinθ-12=0

16Sin²θ-6Sinθ=4

Sinθ(16Sinθ-6)=4

I still can’t work out what to do next.

Doc Al
Mentor
Hint: Let sinθ = x. What kind of equation would you have for x?

Hint: Let sinθ = x. What kind of equation would you have for x?

ax²+bx+c=0

16x²-6x-4=0

Sinθ=6±√-6²-(4×16×-4)∕2×16 (Sorry not sure how to type that so it looks more clear)

Sinθ=0.65
Sinθ=-0.28

Sin–1(0.65)=40.54°
Sin–1(-0.28)=-16.26°

But seeing as it is between 0° and 360°

θ1 = 40.54°
θ2 = 180°-40.54°=139.46° (?)

Hopefully that's right

Ray Vickson
Homework Helper
Dearly Missed

ax²+bx+c=0

16x²-6x-4=0

Sinθ=6±√-6²-(4×16×-4)∕2×16 (Sorry not sure how to type that so it looks more clear)

Sinθ=0.65
Sinθ=-0.28

Sin–1(0.65)=40.54°
Sin–1(-0.28)=-16.26°

But seeing as it is between 0° and 360°

θ1 = 40.54°
θ2 = 180°-40.54°=139.46° (?)
You wrote
$$\sin \theta = 6 \pm \frac{\sqrt{-6^2 - (4 \times 16 \times -4)}}{2} \times 16$$
Did you mean
$$\sin \theta = \frac{6 \pm \sqrt{(-6)^2 - 4\times16\times (-4)}}{2 \times 16}?\:$$
If so, use parentheses, like this:
[6 ± √[(-6)^2 - 4 × 16 ×(-4)]]/(2 × 16).
Better still, use LaTeX.

Anyway, ##x = 0.65## and ##x = -0.28## are nowhere near being solutions of the equation ##16 x^2 - 6 x -4 = 0##. In fact, if ##f(x) = 16 x^2 - 6 x - 4## we have ##f(0.65) = -1.1400## and ##f(-0.28) = -1.0656##.

You wrote
$$\sin \theta = 6 \pm \frac{\sqrt{-6^2 - (4 \times 16 \times -4)}}{2} \times 16$$
Did you mean
$$\sin \theta = \frac{6 \pm \sqrt{(-6)^2 - 4\times16\times (-4)}}{2 \times 16}?\:$$
If so, use parentheses, like this:
[6 ± √[(-6)^2 - 4 × 16 ×(-4)]]/(2 × 16).
Better still, use LaTeX.

Anyway, ##x = 0.65## and ##x = -0.28## are nowhere near being solutions of the equation ##16 x^2 - 6 x -4 = 0##. In fact, if ##f(x) = 16 x^2 - 6 x - 4## we have ##f(0.65) = -1.1400## and ##f(-0.28) = -1.0656##.
Yes I meant that, sorry this is the first time I have tried to type an equation rather than write it.

Not sure what you mean there. It asks for the angles between 0° and 360°, can I not just Sin^-1(x) to get θ?

Doc Al
Mentor
Not sure what you mean there. It asks for the angles between 0° and 360°, can I not just Sin^-1(x) to get θ?
First you must correctly solve the quadratic. Then you can solve for all values of θ that work.

Ray Vickson
Homework Helper
Dearly Missed
Yes I meant that, sorry this is the first time I have tried to type an equation rather than write it.

Not sure what you mean there. It asks for the angles between 0° and 360°, can I not just Sin^-1(x) to get θ?
You obtained ##x=\sin \theta## by solving a quadratic equation; then you used the values of ##x## to get values of ##\theta##. Your values of ##x## are incorrect, so your values of ##\theta## are also incorrect.

Do not take my word for it: test it for yourself. Take your value ##\theta = 40.54^o## and plug it into the function ##f(\theta) = 16 \cos^2(\theta) + 6 \sin(\theta) -12##, to see if you get zero (except for small roundoff errors).

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