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Trigonometrical Equations
normal
સમીકરણ $sin^2 \theta - \frac{4}{{{{\sin }^3}\,\,\theta \,\, - \,\,1}} = 1$$ -\frac{4}{{{{\sin }^3}\,\,\theta \,\, - \,\,1}}$ ને ................ બીજો મળે
A
શૂન્ય
B
એક
C
બે
D
અનંત
Solution
$sin^2 \theta = 1\, [sin\theta \ne \,\, \pm 1 ]$
$\Rightarrow \,sin\theta = 1$
$\Rightarrow \theta = 2n\pi + \pi /2$
$\Rightarrow$ infinite roots
Standard 11
Mathematics
