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Mathematics 26 Online
OpenStudy (anonymous):

WILL GIVE MEDAL AND FAN FOR HELP!!! Can someone please show me step by step how to do this problem?!? Picture attached below.

OpenStudy (anonymous):

OpenStudy (anonymous):

first we take (cosa) as a common factor

OpenStudy (anonymous):

and then we will be having 2 probabilities either cosa = 0 or cosa -1 =0

OpenStudy (anonymous):

and by solving it we can find the values of A

OpenStudy (anonymous):

\[cosA (\cos A -1)=0\]

OpenStudy (anonymous):

did you get it ?

OpenStudy (anonymous):

Else we can also get answer by choice verification since choice was given.. ny verifying choice time eill be saved and useful in competitive exams.

OpenStudy (anonymous):

\[2cos^2(A)-cos(A)=0\]Lets call\[cos(A)=x\]\[2x^2-x=0\]\[x*(2x-1)=0\]Therefore\[x_1=0~~and~~x_2=\frac{1}{2}\]If x=0:\[cos(A)=0~~~~\therefore A=90^o\]If x=1/2 :\[cos(A)=\frac{1}{2}~~~~\therefore A=60^o\]Your answer is:\[S=\{60^o,90^o\}\]

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