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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]
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