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question mark

Sove the equation
(sinx + cosx)1+sin2x = 2
X value ranges between -(pi) & (pi).

Kawal , 10 Years ago
Grade 11
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:Hello student, please find answer to your question
(sinx + cosx)^{1+sin2x} = 2
(sinx + cosx)_{max} = \sqrt{2}
(1+sin2x)_{max} = 2
1+sin2x = 2
sin2x = 1
2x = n\pi + (-1)^{n}\frac{\pi }{2}
x = \frac{n\pi }{2} + (-1)^{n}\frac{\pi }{4}
n = 0, 1
x = \frac{\pi }{4}

sinx + cosx = \sqrt{2}
sin(x+\frac{\pi }{4}) = 1
x + \frac{\pi }{4} = n\pi + (-1)^{n}\frac{\pi }{2}
x = n\pi + (-1)^{n}\frac{\pi }{2}-\frac{\pi }{4}
n = 1
x = \frac{\pi }{4}

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