Posted by **Anonymous** on Monday, December 5, 2011 at 5:27pm.

find all solutions of 2sinx=1-cosx in the interval from 0 to 360

- trig -
**bobpursley**, Monday, December 5, 2011 at 5:32pm
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.

- trig -
**Anonymous**, Monday, December 5, 2011 at 6:18pm
okay so i got -(3/5) and 1, i am supposed to find the solutions in the interval from 0degrees to 360degrees

- trig -
**piyush yadav**, Tuesday, December 6, 2011 at 1:27am
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

- trig -
**piyush yadav**, Tuesday, December 6, 2011 at 1:29am
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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