# sin 2x=cosx

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Subtract both sides by `cosx` to make the right side equal to zero.

`sin2x -cosx = cosx - cosx`

`sin2x - cosx = 0`

` ` We can use the Double - Angle Identity for sine:`sin2x = 2sinxcosx` .

`2sinxcosx - cosx = 0`

` ` Factor out the `cosx` on left side.

`cosx(2sinx - 1) = 0`

Equate each factor to zero.

`cosx =0` and ` ` `2sinx - 1 = 0`

For cosx = 0, we get arccos of both sides.

So, `x = cos^-1(0)` .

==> `x = pi/2` If there, is a given interval like [0, 2pi], **answer will be `pi/2, (3pi)/2` .**

For `2sinx - 1 = 0` .

Add 1 on both sides to isolate the `2sinx`on left side.

`2sinx = 1`

Divide both sides by 2.

`sinx = 1/2`

Getting the arcsin of each sides.

**Answer will be:**

`x = pi/6, (5pi)/6` for interval [0, 2pi]

`x = pi/2, (3pi)/2` for interval [0, 2pi]