Asked by Anonymous
find all solutions of 2sinx=1-cosx in the interval from 0 to 360
Answers
Answered by
bobpursley
square both sides.
4sin^2= 1-2cos+cos^2
4(1-cos^2)=1-2cos+cos^2
now, gather terms, it is a quadratic, solve using the quadratic equation.
4sin^2= 1-2cos+cos^2
4(1-cos^2)=1-2cos+cos^2
now, gather terms, it is a quadratic, solve using the quadratic equation.
Answered by
Anonymous
okay so i got -(3/5) and 1, i am supposed to find the solutions in the interval from 0degrees to 360degrees
Answered by
piyush yadav
the previous ans. is wrong as if we sq. the eqn. it generates false solutions a better method is to divide coefficients of sin and cos by root of sq. of coeff.ofsin+sq. of coeff of cos . that is main pt. remaining we will proceed by making it an argument in cos . and then make general sol. and find no. of sol. or diff solutions by putting value of n in general solution
Answered by
piyush yadav
the previous ans. is wrong as if we sq. the eqn. it generates false solutions a better method is to divide coefficients of sin and cos by root of( sq. of coeff.ofsin+sq. of coeff of cos) . that is main pt. remaining we will proceed by making it an argument in cos . and then make general sol. and find no. of sol. or diff solutions by putting value of n in general solution
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.