How do you find the solution to $sin θ + \sqrt{2} = \frac{\sqrt{2}}{2}$ $sintheta+sqrt2=sqrt2/2$ if $0 \leq θ < 2 π$ $0<=theta<2pi$ ?

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Answer 1 · 2016-12-29T07:26:51

The answer is $S = {\frac{5}{4} π, \frac{7}{4} π}$ $S={5/4pi,7/4pi}$

Let's place $sin θ$ $sintheta$ on one side of the equation

$sin θ + \sqrt{2} = \frac{\sqrt{2}}{2}$ $sintheta+sqrt2=sqrt2/2$

$sin θ = - \sqrt{2} + \frac{\sqrt{2}}{2} = - \frac{\sqrt{2}}{2}$ $sintheta=-sqrt2+sqrt2/2=-sqrt2/2$

$sin θ$ $sintheta$ is $< 0$ $<0$ in the third and fourth quadrant

Therefore,

$θ = \frac{5}{4} π, \frac{7}{4} π$ $theta=5/4pi , 7/4pi$

How do you find the solution to sintheta+sqrt2=sqrt2/2sinθ+√2=√22sintheta+sqrt2=sqrt2/2 if 0<=theta<2pi0≤θ<2π0<=theta<2pi?