SOLVE THE EQUATION SIN2X=2COS2X, FOR 0 DEGREES

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"`2cos^2x+sinx.cosx-sin^2x=0`" ghsprocoach.vn Editorial, 26 July 2012, https://www.ghsprocoach.vn/homework-help/2cos-2x-sinx-cosx-sin-2x-0-350477.Accessed 30 Dec. 2022.

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`2cos^2x + sinxcosx - sin^2x = 0`

To solve for x, consider the above expression as a quadratic equation. Then, factor.

`(2cosx - sinx) ( cosx + sinx) = 0`

Set each factor to zero and solve for x.

> `2cosx - sinx = 0`

Divide both sides by cos x.

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`2 - (sinx)/(cosx) = 0`

 `tanx = 2`

Note that tangent function is positive at quadrants I và III. So,

 `x = 63.43` và `243.43` degrees

The other factor is:

> `cos x + sinx = 0`

Divide both sides by cos x.

`1 + sinx/cosx = 0`

`1 + tanx = 0`

`tan x = -1`

Tangent function is negative at quadrant II and IV. The values of angle x:

`x = 135` and `315` degrees

Hence, the general solution of `2cos^2x + sinxcosx-sin^2x` are:

`x_1 = 63.43 + 360k`degrees

`x_2 = 135 + 360k` degrees

`x_3 = 243.43 + 360k` degrees and

`x_4 = 315 +360k` degrees

 


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