Question

Solve `sin7x+cos2x=-2`?

Answer 1

`x=(2k+1)pi/2" where " k in ZZ`

Explanation:

`sin7x+cos2x=-2`

`=>1+sin7x+1+cos2x=0`

`=>sin^2(7/2x)+cos^2(7/2x)+2sin(7/2x)cos(7/2x)+2cos^2x=0`

`=>(sin(7/2x)+cos(7/2x))^2+2cos^2x=0`

So `sin(7/2x)+cos(7/2x)=0`

`=>tan(7/2x)=-1=tan(-pi/4)`

`=>7/2x=npi-pi/4" where " n in ZZ`

`=>x=(2npi)/7-pi/14" where " n in ZZ`

`=>x=(4n-1)pi/14" where " n in ZZ`
Again
`cosx=0`
`=>x=(2k+1)pi/2" where " k in ZZ`

Please see the modification given below by maganbhi.P

Answer 2

`x=(2k+1)pi/2,kinZ`

Explanation:

Mr.dk_ch had solved the equn. `color(red)(sin7x+cos2x=-2`
`(1)x=(2k+1)pi/2,kinZ` is O.K....But
`color(red)((2)x=(4n-1)pi/14,ninZ ,` is not correct `AA n inZ`
If we take `n=0=>x=-pi/14=>7x=-pi/2and2x=-pi/7`
`LHS=sin(-pi/2)+cos(-pi/7)=-1+cospi/7!=-2`
If we take `n=1=>x=(3pi)/14=>7x=(3pi)/2and2x=(3pi)/7`
`LHS=sin((3pi)/2)+cos((3pi)/7)=-1+cos(3pi)/7!=-2`
If we take `n=2=>x=pi/2=>7x=(7pi)/2and2x=pi`
`LHS=sin((7pi)/2)+cos(pi)=(-1)+(-1)=-2=RHS`
After checking different values of n, I found LHS=RHS only for
`n=...-12,-5,2,9,16,...` which is Arithmetic series.
The `k^(th)` term `t_k=a+(k-1)d,` let,a=9,d=7
`=>t_k=9+(k-1)7=7k+2`
Equn. is satisfies if ,we take
`kinZ` ,for,`n=7k+2` `=>4n=28k+8`
`=>4n-1=28k+7=7(4k+1)`
`=>(4n-1)pi/14=7(4k+1)pi/14=(4k+1)pi/2`
i.,e.`x=(4k+1)pi/2,kinZ`,
which is subset of
`x=(2k+1)pi/2,kinZ`