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5
avatar+122 

Can anyone provide steps to solve these equations?

 

a) 4sinx = cosx (x ranges from -180 to 180), solve with degree

 

b) sin2x = cos(0.5pi - x) (x ranges from 0 to 2pi), solve with radians

 

Thanks!

Mormert  Sep 19, 2018
edited by Mormert  Sep 19, 2018
 #1
avatar+90023 
+2

a) 4sin x  =  cosx        square both sides

 

16sin^2x  =  cos^2x

 

16sin^2x  = 1 -  sin^2x

 

17sin^2x  = 1

 

sin^2x   = 1/17               take both roots

 

sinx  = ±√ (1/17)

 

To find both values in degrees...take the sine inverse

 

arcsin (√[1/17])  ≈  14.04°

 

arcsin (-√ [ 1/17] )  ≈ -14.04°.......however.....we must   subtract this from -180° to get the correct answer

 

-180 - - 14.04  ≈  -165.96°

 

Here's a graph that shows the intersection of these curves : https://www.desmos.com/calculator/c0dyooszqg

 

 

cool cool cool

CPhill  Sep 19, 2018
 #5
avatar
+1

Here is another way:
a) 4sin(x) = cos(x)                     divide both sides by cos(x)
4[sin(x) / cos(x)] =1
4 tan(x) = 1                                divide both sides by 4
tan(x) = 1 / 4
x =arctan(1/4) =~14.04 deg.     Then proceed as CPhill did.

Guest Sep 19, 2018
 #2
avatar+90023 
+2

b)  sin2x  = cos (pi/2  - x)     

 

Note   cos(pi/2 - x)   =  sin x

 

So we have

 

sin2x  = sinx

 

2sinxcosx  =  sinx

 

2sinx cosx  - sinx  = 0

 

sinx ( 2cosx - 1)  = 0

 

So...either   sin x  = 0    and this happens  at    0,  pi  and 2pi

 

Or

 

2cosx - 1  = 0

 

cosx  =  1/2   and this happens at     pi/3   and 5pi/3

 

Here's a graph :  https://www.desmos.com/calculator/y1o4mppidr

 

 

cool cool cool

CPhill  Sep 19, 2018
 #3
avatar+122 
+2

Thank you so much! This really helped me! :D

 

In a) I didn't think of squaring both sides to get cos^2x = 1 -  sin^2x, so thank you!

 

In b) I didn't think of that cos(pi/2 - x)   =  sin x !!! So thank you, this is what helped me the most.

 

Thanks!!! Much love from Sweden.

Mormert  Sep 19, 2018
 #4
avatar+90023 
0

Tack  !!!

 

 

coolcoolcool

CPhill  Sep 19, 2018

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