Solve the following equation for angles between 0 and 360 degrees

4cos²Î¸ + 5sinÎ¸ = 3
4cot² - 6 cosec x = -6

2. As these are trigonometric problems, I have moved this thread accordingly.

Let's look at the first one:

$\displaystyle 4\cos^2(\theta)+5\sin(\theta)=3$

Now, we should observe that if we use a Pythagorean identity:

$\displaystyle \cos^2(\alpha)=1-\sin^2(\alpha)$

We will obtain an equation that is a quadratic in $\sin(\theta)$...

$\displaystyle 4\left(1-\sin^2(\theta)\right)+5\sin(\theta)=3$

Arrange in standard form:

$\displaystyle 4\sin^2(\theta)-5\sin(\theta)-1=0$

Now, what do you get when you apply the quadratic formula?

To be honest I'm lost with it all, I havnt done it in years and I'm hoping to start university soon so was just looking at refresher questions

4. The solutions for $ax^2+bx+c=0$ (a quadratic equation) are

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Using Mark's equation, we have as possible solutions

$$\sin(x)=\frac{-(-5)\pm\sqrt{(-5)^2-4(4)(-1)}}{2(4)}=\frac{5\pm\sqrt{41}}{8}$$

but the sine function has range [-1, 1], so

$$\sin(x)=\frac{5-\sqrt{41}}{8}$$

$$x=\sin^{-1}\left(\frac{5-\sqrt{41}}{8}\right)\approx-10.101^\circ$$

This is a negative so to get solutions in our desired range we use

$$x=360+\sin^{-1}\left(\frac{5-\sqrt{41}}{8}\right)\approx349.899^\circ$$

and

$$x=180-\sin^{-1}\left(\frac{5-\sqrt{41}}{8}\right)\approx190.101^\circ$$

By the way, what do you plan on taking at university? You may need to upgrade your math skills.

this was incredibly helpful, I understand it and just needed my mind refreshing so thanks, in terms of the second one how do you substitute the identity? Is this a given formula or do you need to transpose it to find it yourself? Can you do the same working for the first part of the second question as you did with the first?

6. Originally Posted by simongreen93
this was incredibly helpful, I understand it and just needed my mind refreshing so thanks, in terms of the second one how do you substitute the identity? Is this a given formula or do you need to transpose it to find it yourself? Can you do the same working for the first part of the second question as you did with the first?
$4\cot^2(x) - 6cosec(x) = -6$

For this one the relevant identity is $cot^2(x) = cosec^2(x) - 1 \ \ [\text{eq1}]$ which leads to $4(cosec^2(x)-1) - 6cosec(x) = -6$

$4cosec^2(x) - 4 - 6cosec(x) = -6$

$2cosec^2(x) - 3cosec(x) +1 = 0$ (I've taken the liberty of cancelling out a factor of 2 here)

Just in case you can't remember the identity in [eq1] you can derive it from the Pythagorean formula in post 2.

$\sin^2(x) + \cos^2(x) = 1$

Dividing each side by $\cos^2(x)$ gives the identity required $1 + \cot^2(x) = cosec^2(x)$

7. Originally Posted by simongreen93
... in terms of the second one how do you substitute the identity? Is this a given formula or do you need to transpose it to find it yourself? Can you do the same working for the first part of the second question as you did with the first?
$$4\cot^2(x)-6\csc(x)=-6$$

$$\sin^2(x)+\cos^2(x)=1$$

Divide through by $\sin^2(x)$:

$$1+\cot^2(x)=\csc^2(x)$$

Note that, when dividing, we should always check if there are solutions for when what we are dividing by is equal to 0, but here the original equation is not defined where $\sin(x)=0$ so we're o.k.

So our problem becomes

$$4(\csc^2(x)-1)-6\csc(x)=-6$$

$$4\csc^2(x)-6\csc(x)+2=0$$

$$2\csc^2(x)-3\csc(x)+1=0$$

$$\csc(x)=\frac{3\pm\sqrt{9-8}}{4}=1,\frac12$$

Or one can factor:

$$4\csc^2(x)-6\csc(x)+2=0$$

$$2(2\csc^2(x)-3\csc(x)+1)=0$$

$$(2\csc(x)-1)(\csc(x)-1)=0\implies\csc(x)=1,\frac12$$

$\csc(x)$ has range $(-\infty,-1]\cup[1,\infty)$ so the only (real number) solutions are where $\csc(x)=1$, i.e.,

$$x=\frac\pi2+2k\pi,\,k\in\mathbb{Z}$$

with $x$ in radians.

Note that

$$\arccsc(x)=\arcsin\left(\frac1x\right)$$

$$\arcsec(x)=\arccos\left(\frac1x\right)$$

as problems such as these may not always work out to such 'neat' values.