In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Two acoustic waves are superimposed such that their resulting phase angle ϕ\phiϕ satisfies the equation 6sin(ϕ−60∘)=2cos(ϕ+45∘)\sqrt{6} \sin(\phi - 60^\circ) = 2 \cos(\phi + 45^\circ)6sin(ϕ−60∘)=2cos(ϕ+45∘)
Show that tanϕ=52+3\tan \phi = \frac{5}{2 + \sqrt{3}}tanϕ=2+35 and hence that tanϕ=10−53\tan \phi = 10 - 5\sqrt{3}tanϕ=10−53
Hence or otherwise, solve for 0≤θ<180∘0 \le \theta < 180^\circ0≤θ<180∘, 6sin(3θ−60∘)=2cos(3θ+45∘)\sqrt{6} \sin(3\theta - 60^\circ) = 2 \cos(3\theta + 45^\circ)6sin(3θ−60∘)=2cos(3θ+45∘) giving your answers to one decimal place.