How do you solve $\sqrt{2} sin x - 1 = 0$ $sqrt(2)sinx-1 = 0$ ?

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How do you solve $\sqrt{2} sin x - 1 = 0$ $sqrt(2)sinx-1 = 0$ ?

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Answer 1 · 2018-06-26T10:55:48

General solution: $x = \frac{π}{4} + 2 k π and x = \frac{3 π}{4} + 2 k π$ $x= pi/4 +2 k pi and x = (3 pi)/4+2 k pi$

Explanation:

$\sqrt{2} sin x - 1 = 0 or \sqrt{2} sin x = 1$ $sqrt 2 sin x -1=0 or sqrt 2 sin x =1$ or

$sin x = \frac{1}{\sqrt{2}}; sin (\frac{π}{4}) = \frac{1}{\sqrt{2}}, sin (\frac{3 π}{4}) = \frac{1}{\sqrt{2}},$ $sin x =1/sqrt 2 ; sin (pi/4)= 1/sqrt 2, sin ((3pi)/4)= 1/sqrt 2,$

$:. x= pi/4, x = (3 pi)/4$

Solution: $x= pi/4, x = (3 pi)/4$

General solution: $x= pi/4 +2 k pi and x = (3 pi)/4+2 k pi$

where $k$ is any integer [Ans]

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How do you solve $\sqrt{2} sin x - 1 = 0$ $sqrt(2)sinx-1 = 0$ ?

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How do you solve $\sqrt{2} sin x - 1 = 0$ $sqrt(2)sinx-1 = 0$ ?