Trigonometric Equations - Using Pythagorean Identities | ExamSolutions

[@arbiterb6404]

You could also approach this question by converting the functions so that they are in terms of sin. Firstly, recognise that: tan^2θ = sin^2θ/cos^2θ, also recognise that cosecθ = 1/sinθ = 2cos^2θ / sin^2θ + 8 = 7 / sinθ Next, multiply throughout by sinθ: = 2cos^2θ / sinθ + 8sinθ = 7 Next, multiply throughout by sinθ again: = 2cos^2θ + 8sin^2θ = 7sinθ Recognise that cos^2θ = 1 - sin^2θ = 2(1 - sin^2θ) + 8sin^2θ - 7sinθ = 0 = 6sin^2θ - 7sinθ + 2 = 0 Factorise: (3sinθ - 2)(2sinθ - 1) = 0 Then solve for solutions.

[@S.compEng]

Thank you buddy 👌

[@asaphhere]

Thanks man

[@ExamSolutions_Maths]

@MonkWithAKnife not really, the aim is to get an equation all in one trig. function and you cannot achieve that by making cot the subject..

[@sebastianhay6702]

Hey, Stuart, I wonder how you knew to make everything cosec's when you could have made everything cots...,

[@ExamSolutions_Maths]

How could I have made everything cots? That's why I went for cosec

[@sebastianhay6702]

cosec^2θ=cot^2θ+1. If you square root both sides you get a value for cosecθ in terms of cot! ( cosecθ=cotθ+1)! then Just replace the cosec with 9(cotθ+1)

[@sadmanzaid420]

That doesn't make any sense first of all and the only answer you'd get when cosecθ=cotθ+1 for 0 ≤ θ ≤ 2π is π/2 which is a wrong answer.