Question

cos squared theta minus 3 cos theta + 2 upon sin square theta equal to one

Answer 1

$\frac{ { \cos( \theta ) }^{2} - 3\cos( \theta ) + 2}{ { \sin( \theta ) }^{2} } = 1 \\ \\ { \cos( \theta ) }^{2} - 3\cos( \theta ) + 2 = {\sin( \theta ) }^{2} \\ \\ { \cos( \theta ) }^{2} - 3\cos( \theta ) + 2 =1 - { \cos( \theta ) }^{2} \\ \\ { \cos( \theta ) }^{2} - 3\cos( \theta ) + 2 - 1 + { \cos( \theta ) }^{2} = 0 \\ \\ {2 \cos( \theta ) }^{2} - 3\cos( \theta ) + 1 =0$

${2 \cos( \theta ) }^{2} - 2\cos( \theta ) - \cos( \theta ) + 1 =0 \\ \\ 2 \cos( \theta ) [ \: \cos( \theta ) - 1] - 1[ \: \cos( \theta ) - 1] = 0$

$[\cos( \theta ) - 1]\ [2 \cos( \theta ) - 1]=0\\ \\ 1. \: \cos( \theta ) - 1 = 0 \\ \cos( \theta ) = 1 \\ \cos( \theta ) = \cos( {0}^{o} ) \\ \theta = {0}^{o}$


$2. \: 2 \: \cos( \theta ) - 1 = 0 \\ 2\cos( \theta ) = 1 \\ \cos( \theta ) = \frac{1}{2} \\ \cos( \theta ) = \cos( {60}^{o} ) \\ \theta = {60}^{o}$

$\text{Thus,} \: \boxed{ \theta = 0^o}\ \ \ and\ \ \ \boxed{\theta = 60^o}$

Answer 2

Answer is given below:) :)