Question

If 8sinθcosθcos2θcos4θ=sinx8sinθcosθcos2θcos4θ=sinx  then x=​

Answer 1

Answer:sinx=4+cosx

or, 8sinx−4=cosx

Squaring both side

64sin  

2

x−64sinx+16=cos  

2

x

or, 64sin  

2

x−64sinx+16−1+sin  

2

x=0

or, 65sin  

2

x−64sinx+15=0

or, 65sin  

2

x−39sinx−25sinx+15=0

or, 13sinx(5sinx−3)−5(5sinx−3)=0

or, (13sinx−5)(5sinx−3)=0

either , 13sinx−5=0 OR 5sinx−3=0

or, 13sinx=5 OR 5sinx=3

∴  

sinx=5/13

​

  OR  ∴  

sinx=5/13

​

 

Step-by-step explanation:

Answer 2

Answer:

I do not know

Step-by-step explanation:

I do not know