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# Some trigonometry equations

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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!

Sep 19, 2018
edited by Mormert  Sep 19, 2018

#1
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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   Sep 19, 2018
#5
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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
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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   Sep 19, 2018
#3
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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.

Sep 19, 2018
#4
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Tack  !!!   Sep 19, 2018