## shaqadry 3 years ago 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)

3. mustry

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

4. mustry

$2x=\cos^{-1} 0$

5. mustry

$2\sin2x=1$

6. mustry

$2x=\sin^{-1} 0.5$

7. A.kumar

15 degree

8. ZeHanz

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:$\frac{ 1 }{ 12 }\pi,\frac{ 1 }{ 4 }\pi,\frac{ 5 }{ 12 }\pi,\frac{ 3 }{ 4 }\pi,1\frac{ 1 }{ 12 }\pi,1\frac{ 1 }{ 4 }\pi,1\frac{ 5 }{ 12 }\pi,...$

thank you everyone!

10. ZeHanz

yw!