(sin‌x+cos‌x)1+sin‌2‌x=2, where −π≤x≤π (sin‌x+cos‌x)(sin‌x+cos‌x)2=2[∵(sin‌x+cos‌x)2 =sin2x+cos2x+2‌sin‌cos‌x =1+sin‌2‌x] ∵√a2+b2≤a‌sin‌x+b‌cos‌x≤√a2+b2 ∴−√12+12≤sin‌x+cos‌x≤√12+12 ⇒−√2≤sin‌x+cos‌x≤√2 For −π≤x≤π ∵sin‌x+cos‌x=−√2 at x=−
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sin‌x+cos‌x=√2 at x=
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Now, at x=
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We get (√2)(√2)2=2 =(√2)2=2=2=2 ∴ It is a solution. At x=−
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We get (−√2)(−√2)2=2 (−√2)2=2 ⇒2=2 It is also a solution.