trig
posted by Anonymous .
find all solutions of 2sinx=1cosx in the interval from 0 to 360

square both sides.
4sin^2= 12cos+cos^2
4(1cos^2)=12cos+cos^2
now, gather terms, it is a quadratic, solve using the quadratic equation. 
okay so i got (3/5) and 1, i am supposed to find the solutions in the interval from 0degrees to 360degrees

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

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
Respond to this Question
Similar Questions

maths
hey, i would really appreciate some help solving for x when: sin2x=cosx Use the identity sin 2A = 2sinAcosA so: sin 2x = cos x 2sinxcosx  cosx = 0 cosx(2sinx  1)=0 cosx = 0 or 2sinx=1, yielding sinx=1/2 from cosx=0 and by looking … 
trig
Find all solutions of the equation 2sin^2xcosx=1 in the interval [0,2pi) x1= ? 
Math  Trig
I need to find the solutions in the interval [0, 2pi for: 2sinx + cosx = 0 Im trying to get it where I can split two things up and set them equal to zero but that can only be done when they are being multiplied to each other, but as … 
Trigonometry
Solve for x in the interval 0<=x<360: 1. 2sin2x+cosx =0 2. cos2x=2sinx 3. tanx=2sinx 4. 3cos2x+cosx+2=0 
trig
Solve for [0, 360) 2sinxcosx + cosx =0 2sinxcosx = cosx 2sinx = cosx/cosx sinx = 1/2 {210, 330) Is this correct? 
mathematics
Find all the solutions of 2 sinx=1cosx in the interval from 0° ≤x<360° 
trig
Find all the solutions of 2 sinx=1cosx in the interval from 0° ≤x<360° 
Trig (Last URGENT)
sin2x+cosx=0 , [180,180) = 2sinxcosx+cosx=0 = cosx(2sinx+1)=0 cosx=0 x1=cos^1(0) x1=90 x2=36090 x2=270 270 doesn't fit in [180,180) what do I do? 
trig
Find all solutions in the interval [0,360) sin(4t)=ã3/2 
Trig
find all solutions of 2sinx=12cosx in the interval from 0 to 360