ANSWER:

sin  

2

x+cos  

2

x=1

sin  

4

x+cos  

4

x+2SIN  

2

xcos  

2

x=1

sin  

4

x+cos  

4

x=1−2sin  

2

xcos  

2

x   ...[1]

sin  

2

x+cos  

2

x=1

sin  

6

x+cos  

6

x+3sin  

2

xcos  

2

x(sin  

2

x+cos  

2

x)=1

sin  

6

x+cos  

6

x=1−3sin  

2

xcos  

2

x     ...[2]

∣  

2sin  

4

x+18cos  

2

x

​

−  

2cos  

4

x+18sin  

2

x

​

∣=1

Squaring above equation

∴2sin  

4

x+18cos  

2

x+2cos  

4

x+18sin  

2

x−2(  

2sin  

4

x+18cos  

2

x

​

)(  

2cos  

4

x+18sin  

2

x

​

)=1

∴2(sin  

4

x+cos  

4

x)+18(cos  

2

x+sin  

2

x)−1=2(  

2sin  

4

x+18cos  

2

x

​

)(  

2cos  

4

x+18sin  

2

x

​

)

∴2(1−2sin  

2

xcos  

2

x)+18−1=2(  

2sin  

4

x+18cos  

2

x

​

)(  

2cos  

4

x+18sin  

2

x

​

)       ..( [Using[1] )

∴19−4SIN  

2

xcos  

2

x=2(  

2sin  

4

x+18cos  

2

x

​

)(  

2cos  

4

x+18sin  

2

x

​

)

Squaring above equation

∴361+16sin  

4

xcos  

4

x−152sin  

2

xcos  

2

x=4(4sin  

4

xcos  

4

x+36cos  

6

x+36sin  

6

x+324sin  

2

xcos  

2

x)

∴361+16sin  

4

xcos  

4

x−152sin  

2

xcos  

2

x=16sin  

4

xcos  

4

x+144cos  

6

x+144sin  

6

x+1296sin  

2

xcos  

2

x

∴361=144(cos  

6

x+sin  

6

x)+1448sin  

2

xcos  

2

x

∴361=144(1−3sin  

2

xcos  

2

x)+1448sin  

2

xcos  

2

x    ....( Using [2] )

∴217=1016sin  

2

xcos  

2

x

∴217=254sin  

2

2x

∴sin2x=±  

254

217

​

​

Let sin  

−1

254

217

​

​

=θ

sin2x=±  

254

217

​

​

∴2x=sin  

−1

254

217

​

​

∴2x=θ,π−θ,2π+θ,3π−theta

∴x=  

2

θ

​

,  

2

π−θ

​

,  

2

2π+θ

​

,  

2

3π−θ

​

2x=sin  

−1

(−  

254

217

​

​

)

∴2x=π+θ,2π−θ,3π+θ,4π−θ

∴x=  

2

π+θ

​

,  

2

2π−θ

​

,  

2

3π+θ

​

,  

2

4π−θ

​

So, total NUMBER of solutions =8

So, the answer is option (D)

Step-by-step explanation: