Can anyone solve this, sin2x-tanx=0?

Question

The question is actually in degrees, instead of x it's pheda...I tried solving it but I don't know if it's right, I got 90 degres, 270 degres. Thanks for your help.

Answer 1

sin2x - tanx =
2sinxcosx - sinx/cosx =
sinx(2cosx - 1/cosx) =
0

So sinx = 0, which happens when x = 0,180 or
2cosx = 1/cosx
cosx = +/-(1/root2)
x = 45, 135, 225, 315

Answer 2

pheda? do you mean theta?

Anyway...

sin2X - tanX = 0
2sinXcosX - tanX = 0 (using double angle identity)
2sinXcosX - sinX / cosX = 0
2sinXcos²X - sinX = 0 (multiplying both sides by cosX)
2sinX(1-sin²X) - sinX = 0 (using another identity)
2sinX-2sin³X - sinX = 0
sinX-2sin³X =0 (collecting terms)
sinX(1-2sin²X) =0 (factorising)

sinX=0 or 1-2sin²X = 0

First solution gives X=0, 180, 360 (540, and so on)

Second solution gives 1-2sin²X = 0 so sin²X = 0.5 so sinX = sqrt(0.5) so X= 45, 135, 225, 315

So the possible solutions from 0 to 360 are X = 0, 45, 135, 180, 225, 315, 360

YOU MIGHT WANT TO MAKE THE FONT BIGGER TO VIEW THE ABOVE AS IT IS DIFFICULT TO SEE THE DIFFERENCE BETWEEN 'SQUARED' AND 'CUBED'

Answer 3

If you mean sin^2x - tanx = 0,
sinx(sinx - 1/cosx) = 0
One solution is x = 0
That leaves
sinx - 1/cosx = 0
which has no solution, since the minimum value of 1/cosx is 1, when x = 0, and the maximum value sinx can have is 1

Answer 4

sin2x=2sinxcosx
2sinxcosx - sinx/cosx=0
2sinx cos^2x -sinx=0
sinx(2cos^2x-1)=0
sinx=0 2cos^2x-1=0
: 2cos^2x=1
: cos^2x=1/2
: cosx=+/- (1/√2)
x= 0, 45,135,180, 225, 315,360