\(\displaystyle{18}{{\sec}^{{2}}\theta}-{16}{\tan{\theta}}{\sec{\theta}}-{15}={0}\) Find all solutions in the interval

serpebm2r

serpebm2r

Answered question

2022-03-28

18sec2θ16tanθsecθ15=0 Find all solutions in the interval 0θ<360

Answer & Explanation

Dixie Reed

Dixie Reed

Beginner2022-03-29Added 15 answers

Multiplying throughout by cos2(θ), we get
1816sin(θ)15cos2(θ)=01816sin(θ)15(1sin2(θ))=0
Hence, we get
15sin2(θ)16sin(θ)+3=0
sin(θ)=8±1915
This gives us
θ=2nπ+arcsin(8±1915),(2n+1)πarcsin(8±1915)
i.e,
θ360n+55.48, 360n+14.05, 360n18055.48, 360n+18014.05

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