Question

$3 \sin( \alpha ) = 2 \cos {}^{2} ( \alpha )$
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Answer 1

Step-by-step explanation:

$3 \sin( \alpha ) = 2 { \cos }^{2} \alpha \\ 3 \sin( \alpha ) = 2(1 - { \sin}^{2} \alph = a ) \\ 3 \sin( \alpha ) = 2 - 2 { \sin }^{2} \alpha \\ 3 \sin( \alpha ) + 2 { \sin }^{2} \alpha = 2 \\ 3 \sin( \alpha ) + 2 { \sin }^{2} \alpha - 2 = 0 \\ 2 { \sin }^{2} \alpha + 3 \sin( \alpha ) - 2 = 0 \\ let \: us \: assume \: x = \sin( \alpha ) \\ then \\ 2 {x}^{2} + 3x - 2 = 0 \\ 2 {x}^{2} - x + 4x - 2 = 0 \\ x(2x - 1) + 2(2x - 1) = 0 \\ (2x - 1)(x + 2) = 0 \\ x = ( \frac{1}{2} and \: - 2) \\ we \: assumed \: \ \sin( \alpha ) = x \\ so \: \sin( \alpha ) = ( \frac{1}{2} \: and \: - 2) \: then \: \alpha = 30 \\ \sin( \alpha ) = - 2 \: does \: not \: exist \: so \: the \: answer \: is \: 30 \: degreeee$

sin ø= (1/2)

ø= sin^-1(1/2)=30°