Question

trigonometric equation
.
16. cos x + sin x = 1
find the value of x

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Answer 1

$Given \: \: \: Question \: \: Is \: \\ \\ \cos(x) + \sin(x) = 1 \\ find \: \: \: x \\ \\ Answer \: \\ \\ we \: \: know \: \: that \: \: \\ \sin {}^{2} (x) + \cos {}^{2} (x) = 1 \\ \\ \cos(x) = \sqrt{(1 - \sin {}^{2} (x) )} \\ \\ \sqrt{(1 - \sin {}^{2} (x)) } + \sin(x) = 1 \\ \\ put \: \: \: \sin(x) = z \: \: ...(01)\\ \\ \sqrt{(1 - z {}^{2}) } + z = 1 \\ \\ \sqrt{(1 - z {}^{2} )} = (1 - z) \\ \\ squaring \: \: on \: \: bothsides \: \: \\ \\ (1 - z {}^{2} ) = 1 + z {}^{2} - 2z \\ \\ (1 - z {}^{2} - z {}^{2} - 1 + 2z) = 0 \\ \\ - 2z {}^{2} + 2z = 0 \\ \\ 2z {}^{2} - 2z = 0 \\ \\ 2z(z - 1) = 0 \\ \\ 2z = 0 \: \: \: \: or \: \: \: \: (z - 1) = 0 \\ \\ z = 0 \: \: \: \: or \: \: \: \: z = 1 \\ \\ from \: \: \: \: (01) \\ \\ \sin(x) = 0 \: \: \: \: or \: \: \: \: \sin(x) = 1 \\ \\ \sin(x) = \sin(0) \: \: \: or \: \: \: \: \sin(x) = \sin(90) \\ \\ x = 0 \: \: \: or \: \: \: \: x = 90 \\ \\ therefore \: \: \: \: x = 0 \: \: \: or \: \: \: x = 90$