# Explain what x in the equation 32(sin x)^6+cos6x -1=0?

embizze | High School Teacher | (Level 2) Educator Emeritus

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Solve `32sin^6(x)+cos(6x)-1=0` :

(1) `sin^6(x)=(10-15cos(2x)+6cos(4x)-cos^(6x))/32`

So `32sin^6(x)+cos(6x)-1`

`=10-15cos(2x)+6cos(4x)-cos^6(x)+cos^6(x)-1`

`=9-15cos(2x)+6cos(4x)`

(2) Solve `9-15cos(2x)+6cos(4x)=0`

`3(3-5cos(2x)+2cos(4x))=0`

** `cos(4x)=8cos^4(x)-8cos^2(x)+1` **

`3(3-5cos(2x)+16cos^4(x)-16cos^2(x)+2)=0`

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

`3(5-10cos^2(x)+5+16cos^4(x)-16cos^2(x)+2)=0`

`6(8cos^4(x)-13cos^2(x)+5)=0`

(3) Solve `8cos^4(x)-13cos^2(x)+5=0`

This is quadratic in `cos^2(x)` :

`cos^2(x)=(13+-sqrt(13^2-4(8)(5)))/16`

`cos^2(x)=1 ==> cosx=+-1 ==> x=n pi, n in ZZ`

`cos^2(x)=5/8`

`==>cosx=+-sqrt(5/8)`

`==>x=cos^(-1)sqrt(5/8)+n pi` or `cos^(-1)(-sqrt(5/8))+n pi`

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The solutions are `x=n pi, x~~.6591 + n pi, x~~-.6591 + n pi`

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The graph:

pramodpandey | College Teacher | (Level 3) Valedictorian

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We have

`sin^6(x)+cos(6x)-1=0`

Using De-Moiver's Theorem we can calculate `sin^6(x)`

z=cosx+isinx

z^(-1)=cosx-isinx

z-z^(-1)=2isinx

sinx=(1/(2i))(z-z^(-1))

`sin^6(x)={(1/(2i))(z-z^(-1))}^6` ,Apply binomial theorem on right

side and expand ,apply De-Moiver's Theorem ,we have

`sin^6(x)=(-1/32)(cos(6x)-6cos(4x)+15cos(2x)-10)`

Thus now our problem reduces to

`32 (-1/32)(cos(6x)-6cos(4x)+15cos(2x)-10)+cos(6x)-1=0`

`-cos(6x)+6cos(4x)-15cos(2x)+10+cos(6x)-1=0`

`6cos(4x)-15cos(2x)+9=0`

`3(2cos(4x)-5cos(2x)+3)=0`

`2cos(4x)-5cos(2x)+3=0`

`2(2cos^2(2x)-1)-5cos(2x)+3=0`

`4cos^2(2x)-5cos(2x)+1=0`

`4cos^2(2x)-4cos(2x)-cos(2x)+1=0`

`4cos(2x)(cos(2x)-1)-1(cos(2x)-1)=0`

`` In factorizing above equation ,I have used method of spliting middle

term.

`(4cos(2x)-1)(cos(2x)-1)=0`

Thus either

4cos(2x)-1=0

or

cos(2x)-1=0

If

4cos(2x)-1=0

4cos(2x)=1

cos(2x)=1/4

cos(2x)=cos((21pi)/50)

`2x=2npi+-((21pi)/50)`

`x=npi+-((21pi)/100)`    ,n is an integer.

If

cos(2x)-1=0

cos(2x)=1

`2x=2n_1pi`

`x=n_1pi`  , `n_1`  is an integer.

Here we have used one formula

`cos(2A)=2cos^2(A)-1`