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trig

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

  • trig - ,

    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 - ,

    okay so i got -(3/5) and 1, i am supposed to find the solutions in the interval from 0degrees to 360degrees

  • trig - ,

    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 - ,

    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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