Solve for θ if 2 tan^2 θ + secθ − 4 = 0 0° ≤ θ ≤ 360°

tan^2 θ =sec^2 θ - 1

therefore 2(sec^2θ -1)+sec θ -4=0

2sec^2 θ + sec θ -6 =0
(sec θ + 2) (2sec θ -3)=0

sec θ = 2 or 2secθ =3

this is the furthest i got

2. Originally Posted by joey1
Solve for θ if 2 tan^2 θ + secθ − 4 = 0 0° ≤ θ ≤ 360°

tan^2 θ =sec^2 θ - 1

therefore 2(sec^2θ -1)+sec θ -4=0

2sec^2 θ + sec θ -6 =0
(sec θ + 2) (2sec θ -3)=0

sec θ = 2 or 2secθ =3

this is the furthest i got
$\displaystyle 2\tan^2{\theta} + \sec{\theta} - 4 = 0$.

Now, remembering that $\displaystyle \tan^2{\theta} + 1 = \sec^2{\theta}$, this gives...

$\displaystyle 2(\sec^2{\theta} - 1) + \sec{\theta} - 4 = 0$

$\displaystyle 2\sec^2{\theta} - 2 + \sec{\theta} - 4 = 0$

$\displaystyle 2\sec^2{\theta} + \sec{\theta} - 6 = 0$

$\displaystyle 2\sec^2{\theta} + 4\sec{\theta} - 3\sec{\theta} - 6 = 0$

$\displaystyle 2\sec{\theta}(\sec{\theta} + 2) - 3(\sec{\theta} + 2) = 0$

$\displaystyle (\sec{\theta} + 2)(2\sec{\theta} - 3) = 0$

Case 1:

$\displaystyle \sec{\theta} + 2 = 0$

$\displaystyle \sec{\theta} = -2$

$\displaystyle \cos{\theta} = -\frac{1}{2}$

$\displaystyle \theta = \left\{120^\circ, 240^\circ\right\}$.

Case 2:

$\displaystyle 2\sec{\theta} - 3 = 0$

$\displaystyle 2\sec{\theta} = 3$

$\displaystyle \sec{\theta} = \frac{3}{2}$

$\displaystyle \cos{\theta} = \frac{2}{3}$

$\displaystyle \theta = \left\{\arccos{\frac{2}{3}}, 360^\circ - \arccos{\frac{2}{3}}\right\}$.

3. Originally Posted by joey1
Solve for θ if 2 tan^2 θ + secθ − 4 = 0 0° ≤ θ ≤ 360°

tan^2 θ =sec^2 θ - 1

therefore 2(sec^2θ -1)+sec θ -4=0

2sec^2 θ + sec θ -6 =0
(sec θ + 2) (2sec θ -3)=0

sec θ = - 2 or 2secθ =3

this is the furthest i got
hi good job so far ,

i will use b for convenience

cos b=2/3 , cos b=-1/2 since cos is reciprocal of sec .

cos b=2/3 , cos is positive here so it will be in the 1st and 4th quadrant .

b=48.19 or 311.81

cos b=-1/2 , cos -ve so 2nd and 3rd quadrant .

and we know from the special angles cos 60 =1/2

so b=120 or 240

therefore b=48.19 , 120 , 240 , 311.81

4. Originally Posted by joey1
Solve for θ if 2 tan^2 θ + secθ − 4 = 0 0° ≤ θ ≤ 360°

tan^2 θ =sec^2 θ - 1

therefore 2(sec^2θ -1)+sec θ -4=0

2sec^2 θ + sec θ -6 =0
(sec θ + 2) (2sec θ -3)=0

sec θ = 2 or 2secθ =3

this is the furthest i got