Question

Question #0f638

Answer 1

See below.

Explanation:

This equation is equivalent to

`sqrt(sin^2x) = sqrt(cos^2(3x))`

now squaring both sides

`sin^2x = cos^2(3x) rArr (sinx +cos3x)(sinx-cos3x) = 0`

and now knowing that

`cos3x = cos^3x-3cosx sin^2x`

`{(sinx+cos^3x-3cosx sin^2x = 0),(sinx-cos^3x+3cosx sin^2x = 0):}`

Those equations to be completely solved require the solution of a cubic polynomial. We left this as an exercise.

Answer 2

A different method to the one below.

Explanation:

We have `abs(sinx)=abs(cos3x)`. But by definition, `cos3x=sin(pi/2-3x)`.

So the equation becomes `abs(sinx)=abs(sin(pi/2-3x))`.

This equation is true iff `sinx=sin(pi/2-3x)` or `sinx=-sin(pi/2-3x)=sin(3x-pi/2)`

Finding general solutions will be left as an exercise

Answer 3

`(x=pi/8 + (kpi)/4)`

Explanation:

`|sinx|=|cos3x|`

From the tables we know that :

`color(blue)(cos3x=4 cos^3 x-3 cosx)`,

`|sinx|=|4 cos^3 x-3 cosx|`

`+-sinx= (4cos^2 x-3)cosx`

`+-tan x= (4cos^2 x-3)`

`color(blue)(1+tan^2 x=1/cos^2x)`

`tan^2 x=( (4cos^2 x-3))^2=(16cos^4x-24cos^2 x+9)`

`1/cos^2x=16cos^4x-24cos^2 x+9+1`

`1=16cos^6x-24cos^4 x+10Cos^2x`

I could solve this equation with `y =cos^2x`

`16y^3-24y^2+10y-1=0`

but it is beyond my capabilities.

If we play a little with the trigonometric circle we can find

that `x=+-pi/4+kpi` is one of the solutions, as

`|sin(x)|=|sin(pi/4)+kpi|=sqrt(2)/2`

and

`|cos(3x)|=|cos(3pi/4)+3kpi|=sqrt(2)/2`

`color(red)(x=pi/4+kpi/2`

so `x=pi/4-> cos(x)=sqrt(2)/2-> y=0.5`

We just need to divide the third degree equation by (y-0.5) and obtain the remainder solutions:

`16y^2-16y+2=0`

`8y^2-8y+1=0`

`y=(8+-sqrt(8^2-4*8*1))/(2*8)`

`y=(8+-sqrt(64-32))/(16)`

`y=(8+-sqrt(32))/(16)`

`y=(8+-4sqrt(2))/(16)`

`y=(2+-sqrt(2))/(4)`

`->cos^2x=y`

`cosx=sqrt((2+-sqrt(2))/(4)`

`cosx=sqrt(2+-sqrt(2))/(2)`

`cosx=cos(+-pi/8) or cos x=cos (+-(3pi)/8)`

`x=+-pi/8+2kpi or x +-3pi/8+2kpi`

`color(red)(x=pi/8 + kpi/2)`

Remember the first solution:

`color(red)(x=pi/4+kpi/2`

and gives

`color(green)(x=pi/8 + (kpi)/4)`

Answer 4

x = pi/8 + (kpi)/4
x = - pi/4 + kpi
x = - (3pi/4) + kpi

Explanation:

Note that sin x = cos (pi/2 - x)
`Icos (pi/2 - x)I = Icos 3xI`
We have to solve these equations:
`cos (pi/2 - x) = cos 3x`, and
`cos (pi/2 - x) = - cos 3x`
A. `cos (pi/2 - x) = cos 3x`
`(pi/2 - x) = +- 3x`
a. `pi/2 - x = 3x` --> `4x = pi/2 + 2kpi` --> `x = pi/8 + (kpi)/4`
b. `pi/2 - x = - 3x` --> `2x = - pi/2 + 2kpi` --> `x = - pi/4 + kpi`
B. `cos (pi/2 - x) = - cos 3x = cos (3x + pi)`
`(pi/2 - x) = +- (3x + pi)`
a. `pi/2 - x = 3x + pi` --> `4x = - pi/2 + 2kpi` -->
`x = - pi/4 + kpi`
b. `pi/2 - x = - 3x - pi` --> `2x = - (3pi)/2 + 2kpi` -->
`x = - ((3pi)/4) + kpi`
Finally, the general solutions are:
`x = pi/8 + (kpi)/4; x = - pi/4 + kpi; x = -((3pi)/4) + kpi`
Check.
`x = pi/8 = 22^@5` --> `sin 22.5 = 0.38` -->
`cos 67.5 = 0.38`. Proved
`x = - 45` --> `sin (- 45) = - sqrt2/2` -->
`cos (- 135) = - sqrt2/2`. Proved
`x = - (3pi)/4` --> `sin x = - sqrt2/2` -->
`cos ((9pi)/4) = cos (pi/4) = sqrt2/2`. Proved.