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

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1. Dec 6, 2016

DanRow93

1. The problem statement, all variables and given/known data
Solve acos²θ+bsinθ+c=0 for all values 0≤θ≤360°

a=16
b=6
c=-12

So 16cos²θ+6sinθ-12=0

2. Relevant equations
Cos²x=1-Sin²x

3. 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!

2. Dec 6, 2016

Buffu

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

Where does that three come from ?

3. Dec 6, 2016

DanRow93

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.

4. Dec 6, 2016

DanRow93

16-16Sin²θ+6Sinθ-12=0

16Sin²θ-6Sinθ=4

Sinθ(16Sinθ-6)=4

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

5. Dec 6, 2016

Staff: Mentor

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

6. Dec 6, 2016

DanRow93

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° (?)

7. Dec 6, 2016

DanRow93

Hopefully that's right

8. Dec 6, 2016

9. Dec 6, 2016

Ray Vickson

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 $f(-0.28) = -1.0656$.

10. Dec 6, 2016

DanRow93

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?

11. Dec 6, 2016

Staff: Mentor

First you must correctly solve the quadratic. Then you can solve for all values of θ that work.

12. Dec 6, 2016

Ray Vickson

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).

Last edited: Dec 6, 2016