Math, asked by R2pHas9, 1 year ago

4cos2x+6sinx=6
what is the angle of x?

Answers

Answered by AdityaSharma111

WE KNOW THAT (Sinx)^2=(1-cos2x)÷2--------(1)

FROM HERE, Cos2x can BE WRITTEN AS 1-2sin^2X

NOW WE CAN WRITE 4COS2X+6SINX=6 AS
4(1-2SIN^2X)+6SINX=6
WHICH FURTHER RESOLVES TO 8SIN^2X-6SINX+2=0

LET SINX be "T"

8T^2-6T+2=0,,,,,,,,,,,,,,,

T€R
So there seems no values for the equation

PLEASE MARK IT BRAINLIEST ANSWER.PLEASE

THANKS

AdityaSharma111: Please mark it brainliest answer if your query has been solved.Please :-)

R2pHas9: 4cos^2x should be 4(1-sin^2x). isnt it?

AdityaSharma111: Yes but in question,there was 4cos2X not 4cos^2x thats why

R2pHas9: dude i m having problem writing these equations...My bad!

AdityaSharma111: Is my solution not clear

AdityaSharma111: tell me if you are facing any problem regarding my solution

R2pHas9: thanks for the solution!

AdityaSharma111: :-)

Previous Question

Next Question

4cos2x+6sinx=6what is the angle of x?

Answers

4cos2x+6sinx=6
what is the angle of x?