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.
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
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