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avatar+1124 

Hi friends,

We have two graphs, one for f(x)=Cos2x and another for g(x)=3Sin2x for both of them being between 0 - 180 deg.

 

The question is: Calculate the values for x where f(x) = g(x) if x E (0;180)

 

Please please help me...all help is always much appreciated!..

 Sep 17, 2020
 #1
avatar+14995 
0

We have two graphs, one for f(x)=Cos2x and another for g(x)=3Sin2x

for both of them being between 0 - 180 deg.

The question is: Calculate the values for x where f(x) = g(x) if x \(\in \) (0;180)

 

Hello juriemagic!

 

\(f(x)=g(x)\\ cos\ 2x=3sin\ 2x\)

\(u=2x\)

\(cos\ u=3sin\ u\\ cos^2\ u =3(1-cos^2\ u)\\ cos^2\ u=3-3cos^2\ u\\ 4cos^2\ u=3\\ cos\ u=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\)

\(u=30^{\circ}\\ x=\frac{u}{2}\)

\(x=15^{\circ}\)

laugh  !

 Sep 18, 2020
edited by asinus  Sep 18, 2020
 #2
avatar+1124 
+1

Hi asinus,

the memo I have on this sum uses a way different approach, and comes up with 9,22 or 99,22 deg..?

juriemagic  Sep 18, 2020
 #3
avatar+14995 
+1

 

We have two graphs, one for f(x)=Cos2x and another for g(x)=3Sin2x

for both of them being between 0 - 180 deg.

The question is: Calculate the values for x where f(x) = g(x) if x \(\in \) (0;180)

 

Hello juriemagic!            Answer 1 # is wrong!

 

\(f(x)=g(x)\\ cos\ 2x=3sin\ 2x\)

\(u=2x\)

\(cos\ u=3sin\ u\\ cos^2\ u =9(1-cos^2\ u)\\ cos^2\ u=9-9cos^2\\ 10cos^2\ u=9\\ cos\ u=\pm\sqrt{\frac{9}{10}}=\pm \frac{3}{\sqrt{10}}\)

\(u_1=18.435^{\circ}\\ u_2=161.565^{\circ}\)

\(x=\frac{u}{2}\)

\(x_1=9.217^{\circ}\\ x_2=80.783^{\circ}\)

laugh  !

 Sep 18, 2020

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