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प्रश्न
Find the principal solution and general solution of the following:
sin θ = `-1/sqrt(2)`
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उत्तर
sin θ = `-1/sqrt(2)`
We know that principal of sin θ lies in `[ - pi/2, pi/2]`
sin θ = `- 1/sqrt(2 < 0`
∴ The principal value of sin θ lies in the IV quadrant.
sin θ = `- 1/sqrt(2)`
= `- sin(pi/4)`
sin θ = `sin (- pi/4)`
Hence θ = `- pi/4` is the principal solution.
The general solution is
θ = nπ + (– 1)n . `( pi/4)`, n ∈ Z
θ = `"n"pi + (- 1)^("n" + 1) * pi/4`, n ∈ Z
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