Hello! Please can sombody help me with those three equations?  THANKS

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

`24*cos(x) +5*cos(2x) +2*sin^2(x) +13 =0`

We use the identities:

`cos(2x) =2*cos^2(x) -1`

`sin^2(x) =1-cos^2(x)`

and obtain

`24*cos(x) +5*2*cos^2(x) -5 +2 -2*cos^2(x) +13 =0`

`8*cos^2(x) +24*cos(x) +10 =0`

we substitute `cos(x) =y` and obtain

`4*y^2 +12*y +5=0`

having as solutions

`y_1 =-1/2`  and `y_2 =-5`

Since `|y_2| > 1` it is out of range and we disregard it.

Thus `cos(x) =-1/2` which gives as solutions

`x =pi- pi/3 (+2*n*pi)`  or `x =pi+pi/3 (+2*n*pi)`

Sources:
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valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

Third equation:

c)

`sin(pi/4 -5x) =-sqrt(3)/2`

`pi/4 -5x =pi+pi/3 (+2*n*pi)] or [pi/4 -5x =2pi-pi/3 (+2*n*pi)`  
`-5x =pi+pi/3 -pi/4 (+2*n*pi)`  or ` `

`-5x =2*pi-pi/4-pi/3(+2*n*pi)`

which can be written as

`-5x =13*pi/12 (+2*n*pi) or -5x =5*pi/12(+2*n*pi)`


Finally

`x =-(13/60)*pi (-2/5*n*pi) or x =-pi/12 (-2/5*n*pi)`


(above n is an integer number.)

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nonaaaa | eNotes Newbie

Posted on

How can I right this `-2+i2sqrt(3)`

`TO`

`re^^ itheta`

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