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c2 triq question

Q: For −π⩽Θ⩽π -\pi \leqslant \Theta \leqslant \pi , solve

6sinΘcosΘ−3cosΘ+2sinΘ−1=0 6sin \Theta cos\Theta -3cos\Theta + 2sin\Theta - 1 = 0

my attempt:

6sinΘcosΘ−3cosΘ+2sinΘ−1=0 6sin \Theta cos\Theta -3cos\Theta + 2sin\Theta - 1 = 0
let y=sinΘ y= sin \Theta , x=cosθ x = cos \theta
6yx−3x+2y=1 6yx -3x + 2y = 1
6yx−3x=1−2y 6yx -3x = 1 - 2y
3x(2y−1)=1−2y 3x(2y-1)= 1-2y
3x=1−2y2y−1 3x = \frac{1-2y}{2y-1}
3x=−1 3x =-1
x=−13 x = - \frac{1}{3}

then

cosθ=−13 cos \theta =- \frac{1}{3}

θ=1.91 \theta = 1.91
radians

the full answer when applied to the sketch y=cosθ y = cos \theta −π⩽Θ⩽π -\pi \leqslant \Theta \leqslant \pi ,

the anser i get is θ=−1.91,1.91radians \theta = -1.91, 1.91 radians


but the book gives the answer θ= \theta = π6 \frac{\pi}{6} 5π6\frac{5\pi}{6} , −1.91,1.91radians -1.91, 1.91 radians

please can someone tell me how to find

π6 \frac{\pi}{6}

5π6\frac{5\pi}{6} ,

thank you
Reply 1
Avatar for vc94
vc94
12
When you divide by 2y-1 you're assuming 2y-1 is not zero, so the second case to consider is 2y-1=0 i.e. sin(theta)= 0.5 etc...

Could factorise: 6yx-3x-1+2y=0
(2y-1)(3x+1) =0
Reply 2
Avatar for bigmansouf
bigmansouf
OP
20
Original post by vc94
When you divide by 2y-1 you're assuming 2y-1 is not zero, so the second case to consider is 2y-1=0 i.e. sin(theta)= 0.5 etc...

Could factorise: 6yx-3x-1+2y=0
(2y-1)(3x+1) =0


thank you

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