1) sin( 2*beta ) + 5/9 * cos( beta ) = 0

sin( 2*beta ) = 2*sin( beta )*cos( beta )

sin( 2*beta ) + 5/9 * cos( beta ) = 2*sin( beta )*cos( beta ) + 5/9 * cos( beta ) = 0

cos( beta )*[ 2*sin( beta ) + 5/9 ] = 0

a) cos( beta ) = 0 => beta = pi/2 or beta = 3*pi/2

b) 2*sin( beta ) + 5/9 = 0

sin( beta ) = -5/18 => beta = arcsin( -5/18 ) ..... use calculator

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2) it is more complicated.... probably will solve it later

// i ' ve tried to solve it but it is complcated so the answer is very long and i didnt finished it, probably there is some faster way , some kind of trick...

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3)

the right answer is B) 2

( sin( beta ) + cos( beta ) )^2 + ( sin( beta ) - cos( beta ) )^2 =

= [ sin^2( beta ) + 2*sin( beta )*cos( beta ) + cos^2( beta ) ] + [ sin^2( beta ) - 2*sin( beta )*cos( beta ) + cos^2( beta ) ] =

= 2*( sin^2( beta ) + cos^2( beta ) ) = 2*1 = 2