Question

solve for x if sinx(2cosx+sin60)=cosxcos60 and xE[-180;180]

Answer 1

To solve the equation \( \sin x (2 \cos x + \sin 60^\circ) = \cos x \cos 60^\circ \), we first look at the constants. We know that:

\[
\sin 60^\circ = \frac{\sqrt{3}}{2}
\]
and
\[
\cos 60^\circ = \frac{1}{2}
\]

Substituting these values into the equation gives us:

\[
\sin x (2 \cos x + \frac{\sqrt{3}}{2}) = \cos x \cdot \frac{1}{2}
\]

Now, multiply the right-hand side by 2 to eliminate the fraction:

\[
2 \sin x (2 \cos x + \frac{\sqrt{3}}{2}) = \cos x
\]
\[
2 \sin x (2 \cos x) + \sqrt{3} \sin x = \cos x
\]
\[
4 \sin x \cos x + \sqrt{3} \sin x = \cos x
\]

Rearranging gives:

\[
4 \sin x \cos x + \sqrt{3} \sin x - \cos x = 0
\]

Let’s factor out \(\sin x\):

\[
\sin x (4 \cos x + \sqrt{3}) - \cos x = 0
\]

This gives us two cases to consider:

1. \( \sin x = 0 \)
2. \( 4 \cos x + \sqrt{3} - \frac{\cos x}{\sin x} = 0 \)

**Case 1: \( \sin x = 0 \)**

The solutions for \(\sin x = 0\) within the interval \([-180^\circ, 180^\circ]\) are:

\[
x = 0^\circ, -180^\circ, 180^\circ
\]

**Case 2: Solve \( 4 \cos x + \sqrt{3} - \frac{\cos x}{\sin x} = 0 \)**

This is a bit more complex, so first we can manipulate it again. Rearranging gives:

\[
4 \cos x + \sqrt{3} = \frac{\cos x}{\sin x}
\]
Multiplying both sides by \(\sin x\) (and noting that \(\sin x \neq 0\)) gives:

\[
(4 \cos x + \sqrt{3}) \sin x = \cos x
\]

This looks like a trigonometric identity that might be simpler to solve.

Setting \(U = \frac{\cos x}{\sin x} = \cot x\):

\[
4 \cos x + \sqrt{3} = \cot x
\]

We might have trouble solving this equation algebraically, so revert to simpler boundaries recognizing multiples of angles that equal 0 translates to characteristic \(x\) values.

### Summary

Thus the values for \(x\) so far are:

1. \( x = 0^\circ \)
2. \( x = -180^\circ \)
3. \( x = 180^\circ \)

Therefore, the final solutions for \(x\) in the interval \([-180^\circ, 180^\circ]\) are:

\[
\boxed{0^\circ, -180^\circ, 180^\circ}
\]