Question

#Quality Question

@Trigonometry + Logrithms

In the interval [-π/2,π/2]
the equation

$\bf \large log_{sin \alpha }(cos2 \alpha ) = 2$
has

1) no Solution
2) a Unique Solution
3) two Solution
4) infinite solutions

Full explanation Needed.
​

Answer 1

$\color{brown}{ \large{\qquad \qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  ANSWER \:  \maltese }}}}}}$

$\huge \bf \red{ \mathfrak{b) \: a \: unique \: sol {}^{n} }}$

$\color{maroon}{\qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  \:  \: REQUIRED  \:  \: INFO \:  \:  \:  \maltese }}}}}$

$\bigstar \bf \: \: \: \: log_{n}(m) = x \\ \\ \implies \bf m = {n}^{x} \\ \\ \bigstar \: \: \: \: \bf Range \: of \: sin \theta = [ - 1 ,1 ] \\ \bf when \: \theta \in[ \frac{ - \pi}{2}, \frac{\pi}{2} ]\\ \\ \bigstar \bf \: \: \: \: cos2 \theta = 1 - 2 {sin}^{2} \theta \\ \\ \bigstar \: \: \: \: \bf Base \: of \: Log \: cant't \: be \: Negative$

$\color{magenta} \large{\qquad \underline{ \pmb{{ \mathbb{ \maltese  \:  SOLUTION \:  \maltese }}}}}$

$\huge \mathfrak{ \text{W}e \: \: \text{H}ave}$

$\bf log_{sin \alpha }( \cos2 \alpha ) = 2 \\ \\ \sf cos2 \alpha = ( {sin \alpha })^{2} \\ \\ \sf cos2 \alpha = {sin}^{2} \alpha \\ \\ \sf1 - 2 {sin}^{2} \alpha = sin {}^{2} \alpha \\ \\ \sf 3 {sin}^{2} \alpha = 1 \\ \\ \bf sin \alpha = \pm \frac{1}{ \sqrt{3} } \\ \\ \bf{sin \alpha \ne - ve} \: \: \: ( \because base \: of \: log) \\ \\ \huge \mathfrak{ \text{S}o} \\ \\ \green{ \boxed{ \boxed{ \bf sin \alpha = \frac{1}{ \sqrt{3} } }}}$

$\huge\mathfrak{ \text{H}ence}$

$\large\mathfrak{The \:\: Eq^n \: \: has \: \:a \: \: Unique \: \: Solution}$

Answer 2

Solution:

Starting up with the Given Equation

$\bf log_{sin \alpha }( cos2 \alpha ) = 2 \\ \\ cos2 \alpha = {sin}^{2} \alpha \\(\because log_{y}(x) = p\implies x = {y}^{p})\\ \\ 1 - 2 {sin}^{2} \alpha = sin^{2} \alpha \\(\because cos2\theta=1-2sin^{2}\theta) \\ \\ 3 {sin}^{2} \alpha = 1 \\ \\ sin \alpha = \pm \frac{1}{ \sqrt{3} } \\ \\ sin \alpha \: can't \:be \: negative \\ (\because\:base \: of \: log\: must \: be: positive)\\ \\ \huge\\ \\ \bf \implies sin \alpha = \frac{1}{ \sqrt{3}}$

Hence there is only one solution

option 2 is correct