Question

Solve the equation cos5x = sinx ?

Solve the equation cos5x = sinx , for 0 ≤ x ≤ 360. (Write sinx as cos`(π/2 - x)` ).

Find the general solution of the equation sin3x = cos2x. (Write cos2x as sin`(π/2 - 2x)` )

Answer 1

`x=15^circ,75^circ,135^circ,195^circ,255^circ,315^circor`

`x=67.5^circ,157.5^circ,247.5^circ,337.5^circ`

Explanation:

We know that,

`(1)sintheta= cos(pi/2-theta)`

`color(violet)((2)costheta=cosalpha => theta= 2kpi +-alpha, kinZZ`

Here,

`cos5x = sinx` , where `0 <= x <= 360^circ`

`=>cos5x=cos(pi/2-x) ,where, x in[0, 2pi]`

`=>5x=2kpi+-(pi/2-x),kinZZ...to color(violet)(Apply (2)`

`=>5x=2kpi+pi/2-x or 5x=2kpi-pi/2+x, kinZZ`

We have two options :

`(i)5x=2kpi+pi/2-x, kinZZ`

`=>5x+x=2kpi+pi/2, kinZZ`

`=>6x=(4k+1)pi/2 ,kinZZ`

`=>x=(4k+1)pi/12 ,kinZZ`

`k=0=>color(red)(x=pi/12 and pi/12in [0,2pi]`

`k=1=>color(red)(x=(5pi)/12 and (5pi)/12 in [0, 2pi]`

`k=2=>color(red)(x=(9pi)/12 and (9pi)/12 in [0, 2pi]`

`k=3=>color(red)(x=(13pi)/12 and (13pi)/12 in [0, 2pi]`

`k=4=>color(red)(x=(17pi)/12 and (17pi)/12 in [0, 2pi]`

`k=5=>color(red)(x=(21pi)/12 and (21pi)/12 in [0, 2pi]`

`k=6=>color(blue)(x=(25pi)/12 and (25pi)/12 !in [0, 2pi]`

`(ii)5x=2kpi-pi/2+x, kinZZ`

`=>5x-x=2kpi-pi/2, kinZZ`

`=>4x=(4k-1)pi/2 ,kinZZ`

`=>x=(4k-1)pi/8 ,kinZZ`

`k=0=>color(blue)(x=-pi/8 and -pi/8 !in[0, 2pi]`

`k=1=>color(red)(x=(3pi)/8 and (3pi)/8 in[0,2pi]`

`k=2=>color(red)(x=(7pi)/8 and (7pi)/8 in[0,2pi]`

`k=3=>color(red)(x=(11pi)/8 and (11pi)/8 in[0,2pi]`

`k=4=>color(red)(x=(15pi)/8 and (15pi)/8 in[0,2pi]`

`k=5=>color(blue)(x=(19pi)/8 and (19pi)/8 !in[0,2pi]`

Hence,

`x=pi/12,(5pi)/12,(9pi)/12,(13pi)/12, (17pi)/12, (21pi)/12,(3pi)/8,(7pi)/8,(11pi)/8,(15pi)/8.`

OR

`x=15^circ,75^circ,135^circ,195^circ,255^circ,315^circ,67.5^circ,157.5^circ,247.5^circ,337.5^circ`

Answer 2

We get
`pi/12,3*pi/8,5*pi/12,3*p/4,7pi/8,13pi/12,11pi/8,17pi/12,7/4pi,15/8pi`

Explanation:

Write your equation in the form
`cos(5x)-sin(x)=-2(2sin(2x)-1)*sin(pi/4+x)*sin(pi/4+2x)=0`
and set the single factors equal to Zero.
And for the seond Problem write
`sin(2x)-cos(2x)=-2sin(pi/4-x/2)^2(-1+2cos(2x)-2sin(x))`