Solve the trigonometric equations on the interval 0 le theta < 2pi (32) 4(1 + sin theta) = cos^2 theta (38) sec theta = tan theta + cot theta (40)sqrt(3)sin theta + cos theta = 1

Leila Jennings 2022-07-26 Answered
Solve the trigonometric equations on the interval 0 θ < 2 π.
(32) 4 ( 1 + sin θ ) = cos 2 θ
(38) sec θ = tan θ + cot θ
(40) 3 sin θ + cos θ = 1
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Answers (2)

Osvaldo Crosby
Answered 2022-07-27 Author has 12 answers
(32) Use sin 2 + cos 2 = 1 to write as
4 ( 1 + sin θ ) = 1 sin 2 θ
This is a quadratic equation in sin θ :
sin 2 θ + 4 sin θ + 3 = 0
( sin θ + 2 ) 2 = 1
sin θ = 1 θ = 3 π / 2
(38) Multiply throughout by cos θ :
1 = sin θ + cos 2 θ / sin θ
Multiply throughout by sin θ :
sin θ = sin 2 θ + cos 2 θ
Rewrite using sin 2 + cos 2 = 1
sin θ = 1 θ = π / 2
(40) Divide throughout by cos θ :
3 tan θ + 1 = sec θ
Square both sides:
3 tan 2 θ + 2 3 tan θ + 1 = sec 2 θ
Use sec 2 = 1 + tan 2 :
2 tan 2 θ + 2 3 tan θ = 0
tan θ = 3 θ = π / 3 + n π so here θ = 2 π / 3
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Parker Bird
Answered 2022-07-28 Author has 2 answers
32. 4 ( 1 + sin θ ) = 1 sin 2 θ
so, sin 2 θ + 4 sin θ + 3 = 0
so, ( sin θ + 1 ) ( sin θ + 3 ) = 0
so, sin θ = 1
so, θ = 3 π / 2
38. 1 / cos θ = sin θ / cos θ + cos θ / sin θ
so, sin θ = sin 2 θ + cos 2 θ = 1
so, θ = π / 2
40. 3 sin θ + cos θ = 1
now, divide both sides by 2
so, sin θ cos ( π / 6 ) + sin ( π / 6 ) cos θ = sin ( π / 6 )
so, sin ( θ + π / 6 ) = sin ( π / 6 )
so, θ = 0 , π π / 6 , i e . 0 , 5 π / 6
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