The equation- sin 2 - - 4 sin 2 - -1 -1- 4 sin 2 - -1 has-

The correct option is B no rootdomain: sinθ± 1 after cancelling sin^ 2θ=1 ∴ no roots ...

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General Solution of Trigonometric Equation

The equation:...

Question

The equation:
${sin}^{2} θ - \frac{4}{{sin}^{2} θ - 1} = 1 - \frac{4}{{sin}^{2} θ - 1}$ has:

no root

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one root

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two root

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infinite roots

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Solution

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Similar questions

θ

θ = 0

θ = π / 2

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + s i n^{2} θ & c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & 1 + c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & c o s^{2} θ & 1 + 4 s i n 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

θ

θ = 0

\frac{π}{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & 1 + {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & {cos}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

θ

\frac{π}{2}

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {cos}^{2} θ & {sin}^{2} θ & 4 sin 4 θ {cos}^{2} θ & 1 + {sin}^{2} θ & 4 sin 4 θ {cos}^{2} θ & {sin}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

θ = \frac{a π}{b}

\frac{(b - 3)}{a}

\frac{1}{{sin}^{2} θ} - \frac{1 - {sin}^{2} θ}{1 - {cos}^{2} θ} = 1

\frac{c o s^{2} θ + t a n^{2} θ - 1}{s i n^{2} θ} = t a n^{2} θ

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