Identities, Please Help??

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Answer 1 · 2018-05-05T08:20:10

Please see the explanation below.

$Question (a)$ $"Question (a)"$

$arcsin (x^{2} - 1) = \frac{1}{3} π$ $arcsin(x^2-1)=1/3pi$

$sin (\frac{1}{3} π) = x^{2} - 1$ $sin(1/3pi)=x^2-1$

$x^{2} - 1 = \frac{\sqrt{3}}{2}$ $x^2-1=sqrt3/2$

$x^{2} = (\frac{\sqrt{3}}{2} + 1)$ $x^2=(sqrt 3/2+1)$

$x = \pm \sqrt{(\frac{\sqrt{3}}{2} + 1)} = \pm 1.366$ $x=+-sqrt((sqrt 3/2+1))=+-1.366$

$Question (b)$ $"Question (b)"$

$sin (2 θ + \frac{1}{3} π) = \frac{1}{2}$ $sin(2theta+1/3pi)=1/2$

$(2 θ + \frac{1}{3} π) = arcsin (\frac{1}{2})$ $(2theta+1/3pi)=arcsin(1/2)$

${(2theta+1/3pi=pi/6),(2theta+1/3pi=5/6pi), (2theta+1/3pi=13/6pi),(2theta+1/3pi=17/6pi):}$

$<=>$ , ${(2theta=-1/3pi+pi/6=-1/3pi),(2theta=-1/3pi+5/6pi=pi/2), (2theta=-1/3pi+13/6pi=11/6pi),(2theta=-1/3pi+17/6pi=5/2pi):}$

$<=>$ , ${(theta=-1/6pi),(theta=pi/4), (theta=11/12pi),(theta=5/4pi):}$

The solutions must be for $x in [0, pi]$

The solutions are

$S={pi/4, 11/12pi}$