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1 vote
1 vote
I don’t understand how to use theta (Ø) in trigonometry. I understand what it is, but not how to solve the problems.sin^2(Ø) -4sin(Ø) -5 , where0< (Ø)<360

User Bhavesh N
by
2.4k points

1 Answer

19 votes
19 votes

Given


\sin^2\theta-4\sin\theta-5,0\leq\theta\leq360\degree

Find

Solve for x

Step-by-step explanation

given


\begin{gathered} \sin^2\theta-4\sin\theta-5=0 \\ \sin^2\theta-5\sin\theta+\sin\theta-5=0 \\ \sin\theta(\sin\theta-5)+1(\sin\theta-5)=0 \\ (\sin\theta+1)(\sin\theta-5)=0 \\ so,either \\ (\sin\theta+1)=0,or \\ (\sin\theta-5)=0 \end{gathered}
\sin\theta=5

is not possiblre because


-1\leq\sin\theta\leq1

so , if


\begin{gathered} \sin\theta+1=0 \\ \sin\theta=-1 \\ \theta=\sin^(-1)(-1) \\ \theta=(3\pi)/(2) \end{gathered}

since, general solution is


\theta=(3\pi)/(2)+2n\pi\lbrace n\in Z\rbrace

Final Answer


\theta=(3\pi)/(2)

User Roman Hocke
by
2.9k points
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