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