How do you solve #8\sin ^ { 2} \theta + 17\sin \theta + 9= 0#?

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Answer 1 · 2017-04-27T07:22:33

#theta=-90#

Let #sintheta=a#

So we can write

#8a^2+17a+9=0#

or

#8a^2+8a+9a+9=0#

or

#8a(a+1)+9(a+1)=0#

or

#(a+1)(8a+9)=0#

or

#a+1=0#

or

#a=-1#

or

#sintheta=-1#

or

#theta=-90#--------Ans1

or

#8a+9=0#

or

#8a=-9#

or

#a=-9/8#

or

#sintheta=-9/8#

or

#theta=Invalid#

So we get #theta=-90#