Question

If 270°<m°<360°, and √3sin m° + cos m° = 2sin2011°. then m=?

Answer 1

Question: If $270{}^{\circ} < m{}^{\circ} < 360{}^{\circ}$, and $\sqrt{3}\sin m{}^{\circ} + \cos m{}^{\circ} = 2\sin 2011{}^{\circ}$. then $m = ?$

Solution:

$$\sqrt{3} \sin m{}^{\circ} + \cos m{}^{\circ} = 2 \sin 2011{}^{\circ}$$

Let's divide both sides of the equation (1) by 2:

$$\frac{\sqrt{3}}{2} \sin m{}^{\circ} + \frac{1}{2} \cos m{}^{\circ} = \sin 2011{}^{\circ}$$

Let's substitute $\frac{\sqrt{3}}{2}$ with $\cos 30{}^{\circ}$ and $\frac{1}{2}$ with $\sin 30{}^{\circ}$:

$$\cos 30{}^{\circ} \sin m{}^{\circ} + \sin 30{}^{\circ} \cos m{}^{\circ} = \sin 2011{}^{\circ}$$

Left side of the equation is sinus of the sum of $30{}^{\circ}$ and $m{}^{\circ}$.

$$\sin(m + 30{}^{\circ}) = \sin 2011{}^{\circ}$$ $$\sin(m + 30{}^{\circ}) - \sin 2011{}^{\circ} = 0$$

Here let's use the formula for the subtraction of sinuses: $\sin a - \sin b = 2 \sin \frac{a - b}{2} \cos \frac{a + b}{2}$

$$2 \sin \frac{m + 30 - 2011}{2} \cos \frac{m + 30 + 2011}{2} = 0$$ $$\sin \frac{m - 1981}{2} \cos \frac{m + 2041}{2} = 0$$

From equation (2) we get these solutions: $\sin \frac{m - 1981}{2} = 0$ or $\cos \frac{m + 2041}{2} = 0$.

1) $\sin \frac{m - 1981}{2} = 0$

$$\frac{m - 1981}{2} = \pi n$$ $$m - 1981 = 2 \pi n$$ $$m = 1981 + 2 \pi n$$

Here $1981 = 181{}^{\circ} + 5 \cdot 2\pi n$, so the equation (3) becomes:

$m = 181{}^{\circ} + 2\pi n$, where $n$ is an integer.

This solution does not correspond to the condition that $270{}^{\circ} < m{}^{\circ} < 360{}^{\circ}$.

2) $\cos \frac{m + 2041}{2} = 0$

$$\frac{m + 2041}{2} = \frac{\pi}{2} + \pi k$$ $$m + 2041 = \pi + 2\pi k$$

As $2041 = 61{}^{\circ} + \pi + 5 \cdot 2\pi k$, the equation (4) becomes:

$m = -61{}^{\circ} + 2\pi k$, where $k$ is an integer.

In the last equation substituting $k = 1$ we get a solution $m = 299{}^{\circ}$ that corresponds to the condition that $270{}^{\circ} < m{}^{\circ} < 360{}^{\circ}$.

Answer: $m = 299{}^{\circ}$