Trigonometric EquationHard
Question
Let θ, φ, ∈ [0, 2π] be such that 2cos θ (1 - sinφ) = sin2 θ 
cos φ - 1, tan(2 π - θ) > 0 and - 1 < sin θ < - 
Then φ cannot satisfy
cos φ - 1, tan(2 π - θ) > 0 and - 1 < sin θ < - Then φ cannot satisfyOptions
A.
B.0 < φ < 
C.
D.
< φ < 2π
< φ < 2π Solution
0 cos θ (1 - sin φ) = 
cosφ- 1 = 2 sin θ cos φ - 1
2 cos θ - 2 cos θ sin φ = 2 sin θ cosφ - 1
2 cos θ + 1 = 2 sin (θ + φ)
tan(π - θ) > 0 ⇒ tan θ < 0 and - 1 < sin θ < -
⇒ θ ∈
< sin (θ + φ) < 1
⇒ 2π +
< θ + φ < 
+ 2π
2π +
-θmax < φ < 2π + 
-θmin
cosφ- 1 = 2 sin θ cos φ - 12 cos θ - 2 cos θ sin φ = 2 sin θ cosφ - 1
2 cos θ + 1 = 2 sin (θ + φ)
tan(π - θ) > 0 ⇒ tan θ < 0 and - 1 < sin θ < -
⇒ θ ∈
< sin (θ + φ) < 1 ⇒ 2π +
< θ + φ < + 2π 2π +
-θmax < φ < 2π + -θminCreate a free account to view solution
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