smc-6412-25-Quadratics

Trigonometry, 2ADV T2 EQ-Bank 25

Express `5cot^2 x-2text(cosec)\ x + 2` in terms of `text(cosec)\ x` and hence solve

`5cot^2 x-2text(cosec)\ x + 2 = 0` for `0 < x < 2pi`. (3 marks)

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`x = pi/2`

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`cot^2 x= (cos^2 x)/(sin^2 x)= (1-sin^2 x)/(sin^2 x)= text(cosec)^2 x-1`

`5cot^2 x-2text(cosec)\ x + 2`	`= 0`
`5(text(cosec)^2 x-1)-2text(cosec)\ x + 2`	`= 0`
`5text(cosec)^2 x-2text(cosec)\ x-3`	`= 0`
`(5text(cosec)\ x + 3)(text(cosec)\ x-1)`	`= 0`

`text(cosec)\ x`	`= -3/5`	`text(cosec)\ x`	`= 1`
`sinx`	`= -5/3`	`sinx`	`= 1`
`(text(no solution))`		`x`	`= pi/2`

`:. x = pi/2`

Trigonometry, 2ADV T2 2014 HSC 15a

Find all solutions of `2 sin^2 x + cos x-2 = 0`, where `0 <= x <= 2pi`. (3 marks)

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`x = pi/3,\ pi/2,\ (3pi)/2,\ (5pi)/3`

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♦ Mean mark 42%

`2 sin^2 x + cos x-2`	`= 0`
`2(1-cos^2x) + cos x-2`	`= 0`
`2-2cos^2x + cosx-2`	`= 0`
`-2cos^2x + cosx`	`= 0`
`cosx (-2 cosx + 1)`	`= 0`

`:. -2 cosx + 1`	`= 0`	`\ text(or)\ \ \ \ \ \ \ `	`cos x`	`= 0`
`2 cos x`	`= 1`		`x`	`= pi/2,\ (3pi)/2`
`cos x`	`= 1/2`
`cos(pi/3)`	`=1/2`

`text(S)text(ince cos is positive in)\ 1^text(st) // 4^text(th)\ text(quadrants:)`

`x= pi/3,\ 2 pi-pi/3= pi/3,\ (5pi)/3`

`:. x = pi/3,\ pi/2,\ (3pi)/2,\ (5pi)/3\ \ text(for)\ \ 0 <= x <= 2pi`