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solve \\( \\sin ( x ) = 0.23 \\) on \\( 0 \\leq x < 2 \\pi \\). there a…

Question

solve \\( \sin ( x ) = 0.23 \\) on \\( 0 \leq x < 2 \pi \\).
there are two solutions, a and b, with \\( a < b \\).
\\( a = \\)
\\( b = \\)
give your answers accurate to 3 decimal places.
question help: video

Explanation:

Step1: Find the reference angle

Since \(\sin(x) = 0.23\), the reference angle \(x_{ref}=\arcsin(0.23)\).
Using a calculator, \(x_{ref}\approx0.232\) (in radians).

Step2: Find the first - quadrant solution (A)

For \(0\leq x<2\pi\), the first - quadrant solution \(A = x_{ref}\). So \(A=\arcsin(0.23)\approx0.232\).

Step3: Find the second - quadrant solution (B)

The sine function is positive in the first and second quadrants. The formula for the second - quadrant solution of \(\sin(x)=k\) (\(k>0\)) is \(x=\pi - x_{ref}\).
So \(B=\pi-\arcsin(0.23)\).
Substitute \(x_{ref}\approx0.232\) into the formula: \(B\approx3.142 - 0.232=2.910\).

Answer:

\(A = 0.232\), \(B = 2.910\)