+0

# Cos2x and 3Sin2x

+1
57
3
+748

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)

Sep 17, 2020

#1
+10319
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!

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

$$u=2x$$

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

$$u=30^{\circ}\\ x=\frac{u}{2}$$

$$x=15^{\circ}$$

!

Sep 18, 2020
edited by asinus  Sep 18, 2020
#2
+748
+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
+10319
+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!

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

$$u=2x$$

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

$$u_1=18.435^{\circ}\\ u_2=161.565^{\circ}$$

$$x=\frac{u}{2}$$

$$x_1=9.217^{\circ}\\ x_2=80.783^{\circ}$$

!

Sep 18, 2020