1.

Find the number of solution of the equation `sqrt(cos 2x+2)=(sin x + cos x)` in `[0, pi]`.

Answer» We have `sqrt(cos 2x+2)=(sin x + cos x)`
Squaring both sides, we get
`2+cos 2x=1 + sin 2x`
`:. Sin 2x-cos 2x =1`
Again, squaring both sides, we get
`1-sin 4x =1`
`:. Sin 4x =0`
`rArr 4x=n pi, n in Z`
or `x=(npi)/4`, where `x in [0, pi]`
`:. x=0, pi/4, pi/2, (3pi)/4, pi`
But only `x=pi/4` and `pi/2` satisfy the original equation.
2. Never cancel the terms containing unknown terms which are in product on the two sides. It may cause the loss of a genuine solution.


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