How do you solve $9 + {sin}^{2} x = 10$ $9+sin^2x=10$ for $0 \leq x \leq 2 π$ $0<=x<=2pi$ ?

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Answer 1 · 2018-03-11T18:25:13

$\frac{π}{2}, \frac{3 π}{2}$ $color(blue)(pi/2 , (3pi)/2)$

$9 + {sin}^{2} x = 10$ $9+sin^2x=10$

Subtract 9 from both sides:

${sin}^{2} x = 1$ $sin^2x=1$

$sin x = \pm \sqrt{1} = \pm 1$ $sinx=+-sqrt(1)=+-1$

$x = arcsin (sin x) = arcsin (1) \Rightarrow x = \frac{π}{2}$ $x=arcsin(sinx)=arcsin(1)=>x=color(blue)(pi/2)$

$x = arcsin (sin x) = arcsin (- 1) \Rightarrow x = \frac{3 π}{2}$ $x=arcsin(sinx)=arcsin(-1)=>x=color(blue)((3pi)/2 )$

How do you solve 9+sin^2x=109+sin2x=109+sin^2x=10 for 0<=x<=2pi0≤x≤2π0<=x<=2pi?