Question #23b80

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Answer 1 · 2018-01-26T12:53:10

$θ = 60^{o} or θ = 300^{o}$ $theta= 60^o or theta= 300^o$

$2cos^2θ−3sinθ=0$

$2(1-sin^2theta)−3sinθ=0$ because $cos^2θ+sin^2theta=1$

$2-2sin^2theta−3sinθ=0$

$2sin^2theta+3sinθ-2=0$

$sinθ=(-3+-sqrt(3^2-4*2*(-2)))/(2*2)$

$sinθ=(-3+-sqrt(9+16))/4$

$sinθ=(-3+-sqrt(25))/4=(-3+-5)/4$

$sin theta=-8/4 or sin theta= 2/4$

$sin theta=-2 or sin theta= 1/2$

$theta notin RR or sin theta= 1/2$

$theta= 60^o or theta= (-60^o +360^o)$

$theta= 60^o or theta= 300^o$