1×3tanx+1×cotx=5cosecx
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Hello!
3 tanx + cotx = 5 cosecx
⇒ 3 (sinx)² + (cos)² = 5 cosx
⇒ 3 sin²x + cos²x = 5 cosx
∵ sin²x = 1 - cos² x
⇒ 3 - 3 cos²x - cos²x = 5 cosx
Let " cosx " = t :
3 - 3t² - t² = 5t
⇒ 2t² + 5t - 3 = 0
⇒ (2t - 1).(t + 3) = 0
∴ t = 1 / 2 or t = - 3
Since, t = cosx > - 1
∴ - 3 is not included.
cosx = 1 / 2 ⇒ x = 60°
Cheers!
3 tanx + cotx = 5 cosecx
⇒ 3 (sinx)² + (cos)² = 5 cosx
⇒ 3 sin²x + cos²x = 5 cosx
∵ sin²x = 1 - cos² x
⇒ 3 - 3 cos²x - cos²x = 5 cosx
Let " cosx " = t :
3 - 3t² - t² = 5t
⇒ 2t² + 5t - 3 = 0
⇒ (2t - 1).(t + 3) = 0
∴ t = 1 / 2 or t = - 3
Since, t = cosx > - 1
∴ - 3 is not included.
cosx = 1 / 2 ⇒ x = 60°
Cheers!
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