How do you solve the equation $sqrt2sinx-1=0$ for $-360^circ<x<=360^circ$ ?

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How do you solve the equation $sqrt2sinx-1=0$ for $-360^circ<x<=360^circ$ ?

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Answer 1 · 2018-01-09T13:31:18

$- 225^@; - 315^@; 45^@; 135^@$

Explanation:

$sqrt2sin x - 1 = 0$
$sin x = 1/sqrt2 = sqrt2/2$
Trig table and unit circle give 2 solutions:
$x = pi/4, or 45^@$ and $x = (3pi)/4, or 135^@$
Use the unit circle to select the answers -->
In the period ( - 360 , 0) there are 2 answers:
$x = - 225^@$ , and $x = - 315^@$
In the period (0, 360), there are 2 answers:
$x = 45^@$ , and $x = 135^@$

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How do you solve the equation $sqrt2sinx-1=0$ for $-360^circ<x<=360^circ$ ?

1 Answer

Explanation:

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How do you solve the equation $sqrt2sinx-1=0$ for $-360^circ<x<=360^circ$ ?