Here's the question you clicked on:
shaqadry
solve sin 4x - cos 2x = 0
sin4x-cos2x=0 2sin2xcos2x-cos2x=0 2sinucosu-cosu=0 (2sinu-1)cosu=0 2sinu-1=0 or cosu=0 2sin=1 u=(pi/2)+n pi sinu=1/2 u=(pi/6)+2npi or u=(5pi/6)+2npi replace u with 2x ,then solve to x
x=(pi/12)+n pi x=(5pi/12)+n pi x=(pi/4)+n(pi/2)
sin4x-cos2x=0 sin(2x+2x)-cos2x=0 (sin2xcos2x+cos2xsin2x)-cos2x=0 2sin2xcos2x-cos2x=0 cos2x(2sin2x-1)=0 cos2x=0 or 2sin2x-1=0
This is my try of the problem: I'm going to use a well-known formula for double angles:\[\sin2x=2 \sin x \cos x\]You can use this formula to reduce the sine of 2x to sines and cosines of half of 2x (=x). This means e.g. :\[\sin50x=2 \sin 25x \cos 25 x\]so again: only half of the original 50x is left. Now apply it to the equation:\[\sin 4x - \cos 2x= 0\]then becomes \[2\sin2x \cos 2x -\cos 2x =0\]Factor out the cos 2x:\[\cos 2x \cdot (2\sin2x-1)=0\]This makes everything much easier, because this is an equation of the form: \[a \cdot b = 0\]YOu can split it up into:\[a=0 \vee b=0\]So here that is:\[\cos 2x=0 \vee 2\sin 2x -1 =0\]\[\cos 2x = 0 \vee \sin 2x = \frac{ 1 }{ 2 }\]Now it is all standard work to be done from here:\[\cos 2x=0 \Leftrightarrow 2x=\frac{ 1 }{ 2 } \pi + k \pi \Leftrightarrow x = \frac{ 1 }{ 4 }\pi + \frac{ 1 }{ 2 }k \pi\]\[\sin 2x = \frac{ 1 }{ 2 } \Leftrightarrow 2x=\frac{ 1 }{ 6 }\pi + 2k \pi \vee 2x= \frac{ 5 }{ 6 }\pi + 2k \pi\]\[x=\frac{ 1 }{ 12 }\pi + k \pi \vee x = \frac{ 5 }{ 12 } \pi + k \pi\] Everytime there is a "k" in the anser, it means an integer. If you want to understand better what kind of numbers all these x-values are, you can also list them as follows:\[\frac{ 1 }{ 12 }\pi,\frac{ 1 }{ 4 }\pi,\frac{ 5 }{ 12 }\pi,\frac{ 3 }{ 4 }\pi,1\frac{ 1 }{ 12 }\pi,1\frac{ 1 }{ 4 }\pi,1\frac{ 5 }{ 12 }\pi,...\]