Trigonometric Equation Solutions 5
Find all angles $ x $, $ 0^\circ \le x < 180^\circ, $ such that
\[\sin 6x + \cos 4x = 0.\]Enter all the solutions, separated by commas.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
We can write
\[\sin 6x + \cos 4x = \sin 6x + \sin (90^\circ - 4x).\]Then from the sum-to-product formula,
\begin{align*}
\sin 6x + \sin (90^\circ - 4x) &= 2 \sin \left( \frac{6x + 90^\circ - 4x}{2} \right) \cos \left( \frac{6x - (90^\circ - 4x)}{2} \right) \\
&= 2 \sin (x + 45^\circ) \cos (5x - 45^\circ).\end{align*}Thus, $ \sin (x + 45^\circ) = 0 $ or $ \cos (5x - 45^\circ) = 0 $.
If $ \sin (x + 45^\circ) = 0, $ then $ x = 135^\circ $.
If $ \cos (5x - 45^\circ) = 0, $ then $ 5x - 45^\circ $ must be $ 90^\circ, $ $ 270^\circ, $ $ 450^\circ, $ $ 630^\circ, $ or $ 810^\circ $. These lead to the solutions $ \boxed{27^\circ, 63^\circ, 99^\circ, 135^\circ, 171^\circ} $.
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