How to solve 12sinx-1.8cosx=10 without a graphing calculator?

I tried squaring both sides but can't seem to simplify it. Any ideas on how this can be solved?

Thanks!

Edit: For those who are interested, I have attached the question that has prompted me to solve this system of equations.

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ehild
Homework Helper
Both sinx and cosx can be expressed by tan(x/2)

$$\sin(x)=\frac{2\tan(x/2)}{1+\tan^2(x/2)}$$

$$\cos(x)=\frac{1-\tan^2(x/2)}{1+\tan^2(x/2)}$$

With these substitutions, you get a quadratic equation for tan(x/2).

ehild

vela
Staff Emeritus
Homework Helper
When you have a combination like ##A \sin\theta + B \cos\theta##, a useful trick is to write
$$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}\left(\frac{A}{\sqrt{A^2+B^2}} \sin\theta + \frac{B}{\sqrt{A^2+B^2}}\cos\theta\right)$$Now identify
\begin{align*}
\cos\phi &= \frac{A}{\sqrt{A^2+B^2}} \\
\sin\phi &= \frac{B}{\sqrt{A^2+B^2}}
\end{align*}then you have
$$A \sin\theta + B \cos\theta = \sqrt{A^2+B^2}(\cos\phi \sin\theta + \sin\phi\cos\theta) = \sqrt{A^2+B^2}\sin(\theta+\phi)$$where ##\tan\phi = B/A##.

In this particular problem, if you move the factor of 4 kN over to the righthand side, you have
$$3 \sin \theta - 0.45 \cos \theta = 2.5$$and ##\tan\phi = -0.45/3##, from which you should recognize ɸ as being the angle the line connecting A to the point where the force acts makes with the horizontal.

SammyS
Staff Emeritus
Homework Helper
Gold Member
I tried squaring both sides but can't seem to simplify it. Any ideas on how this can be solved?

Thanks!

Edit: For those who are interested, I have attached the question that has prompted me to solve this system of equations.
What system of equations.

I see none.

ehild
Homework Helper
What system of equations.

I see none.

It is in the title. 12sinx-1.8cosx=10

ehild

SammyS
Staff Emeritus