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I) Solve the equation sin2x + 3cos2x = 0 (for 0 till 360°) ii) How many solutions has the equation sin2x + 3cos2x =0 for 0 <= X <=1080°?

I) Solve the equation sin2x + 3cos2x = 0 (for 0 till 360°) ii) How many solutions has the equation sin2x + 3cos2x =0 for 0 <= X <=1080°?
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I) Solve the equation sin2x + 3cos2x = 0 (for 0 till 360°)

ii) How many solutions has the equation sin2x + 3cos2x =0 for 0 <= X <=1080°?

asked
User Krishna Prasad
by
8.1k points

1 Answer

1 vote
Get rid of cos2x by dividing both the values. So Sin2x/cos2x +3cos2x/cos2x.
Tan2x = 3
2x = -71.5 so x is -35.6
Use the quadrant method and add 360 twho the two values tou get.
answered
User Bjoern
by
8.5k points
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