# Thread: Sine and Cosine question

1. ## Sine and Cosine question

I am learning sine, cosine and trigonometry graphs, and other trig stuff. And I am really stuck with this question:

1) What is the value of x???
2sinx + cosx = 1.5 (x is between 0 and 180)

I would really appreciate it if someone could help me with this question.
Thank You ^^

2. Originally Posted by Mathsaholic
I am learning sine, cosine and trigonometry graphs, and other trig stuff. And I am really stuck with this question:

1) What is the value of x???
2sinx + cosx = 1.5 (x is between 0 and 180)

I would really appreciate it if someone could help me with this question.
Thank You ^^
If you have to solve an equation of the form $a\sin x+b\cos x=c$, you can divide the whole equation by $\sqrt{a^2+b^2}$. The new coefficients of $\sin x$ and $\cos x$ have the property that the sum of their squares is 1. This allows you to apply the Addition Formula for the $\sin$, which says that $\sin(x)=\sin(x)\cos(x_0)+\cos(x)\sin(x_0)$.
Then you use the two equations $\cos(x_0)=a/\sqrt{a^2+b^2}$ and $\sin(x_0)=b/\sqrt{a^2+b^2}$, to figure out what $x_0$ is.

So in your case this works as follows
$2\sin(x)+\cos(x)=1.5$
$\frac{2}{\sqrt{5}}\sin(x)+\frac{1}{\sqrt{5}}\cos(x )=\frac{1.5}{\sqrt{5}}$
$\sin(x+x_0)=\frac{1.5}{\sqrt{5}}$
Now you apply $\sin^{-1}$ to both sides to get the general solution
$x+x_0\approx \begin{cases}42.13^\circ+n\cdot360^\circ\\137.87^\ circ+n\cdot 360^\circ\end{cases}, n\in \mathbb{Z}$

Finally you have that $\cos(x_0)=\frac{2}{\sqrt{5}}$ and $\sin(x_0)=\frac{1}{\sqrt{5}}$, in other words $\tan(x_0)=\sin(x_0)/\cos(x_0)=\frac{1}{2}$, which means, $x_0\approx 26.57^\circ$ (we don't need the general solution for $x_0$ here).

Thus you have

$x\approx \begin{cases}42.13^\circ-26.57^\circ+n\cdot360^\circ\\137.87^\circ-26.57^\circ+n\cdot 360^\circ\end{cases}, n\in \mathbb{Z}$
Now, from these infinitely many solutions you need to pick those, that lie in the reqired intervall.