Quadratic-type and reciprocal equations questions with answers

Questions

1. Question 1Quadratic in sin x, cos x or tan x

   Solve $4\cos^2 x - 3 = 0$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: Let y = cos x: it's a quadratic in y$4y^2 - 3 = 0$
   2. Step 2: Solve for y (two square roots)$4y^2 = 3$$y^2 = \dfrac{3}{4}$$y = \pm\dfrac{\sqrt{3}}{2}$
   3. Step 3: Basic angle: use the positive value$\alpha = \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$$= 30^\circ$
   4. Step 4: ASTC (All, Sine, Tan, Cos): in which quadrants does cos have this sign?$\cos x$ is positive in quadrants 1 and 4.
   5. Step 5: List every angle x in the range, quadrant by quadrant$x = 30^\circ$$x = 360^\circ - 30^\circ$$= 330^\circ$
   6. Step 6: ASTC (All, Sine, Tan, Cos): in which quadrants does cos have this sign?$\cos x$ is negative in quadrants 2 and 3.
   7. Step 7: List every angle x in the range, quadrant by quadrant$x = 180^\circ - 30^\circ$$= 150^\circ$$x = 180^\circ + 30^\circ$$= 210^\circ$
2. Question 2Quadratic in sin x, cos x or tan x

   Solve $4\sin^2 x - 1 = 0$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: Let y = sin x: it's a quadratic in y$4y^2 - 1 = 0$
   2. Step 2: Solve for y (two square roots)$4y^2 = 1$$y^2 = \dfrac{1}{4}$$y = \pm\dfrac{1}{2}$
   3. Step 3: Basic angle: use the positive value$\alpha = \sin^{-1}\left(\dfrac{1}{2}\right)$$= 30^\circ$
   4. Step 4: ASTC (All, Sine, Tan, Cos): in which quadrants does sin have this sign?$\sin x$ is positive in quadrants 1 and 2.
   5. Step 5: List every angle x in the range, quadrant by quadrant$x = 30^\circ$$x = 180^\circ - 30^\circ$$= 150^\circ$
   6. Step 6: ASTC (All, Sine, Tan, Cos): in which quadrants does sin have this sign?$\sin x$ is negative in quadrants 3 and 4.
   7. Step 7: List every angle x in the range, quadrant by quadrant$x = 180^\circ + 30^\circ$$= 210^\circ$$x = 360^\circ - 30^\circ$$= 330^\circ$
3. Question 3sec, cosec and cot

   Solve $\cot x = -\dfrac{\sqrt{3}}{3}$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: cot x = 1 ÷ tan x: turn it round$\tan x = \dfrac{1}{-\dfrac{\sqrt{3}}{3}}$$= -\sqrt{3}$
   2. Step 2: Basic angle: use the positive value$\alpha = \tan^{-1}\left(\sqrt{3}\right)$$= 60^\circ$
   3. Step 3: ASTC (All, Sine, Tan, Cos): in which quadrants does tan have this sign?$\tan x$ is negative in quadrants 2 and 4.
   4. Step 4: List every angle x in the range, quadrant by quadrant$x = 180^\circ - 60^\circ$$= 120^\circ$$x = 360^\circ - 60^\circ$$= 300^\circ$
4. Question 4sec, cosec and cot

   Solve $\sec x = -\dfrac{2\sqrt{3}}{3}$ for $0^\circ \leq x \leq 360^\circ$.

   Show the worked solution

   1. Step 1: sec x = 1 ÷ cos x: turn it round$\cos x = \dfrac{1}{-\dfrac{2\sqrt{3}}{3}}$$= -\dfrac{\sqrt{3}}{2}$
   2. Step 2: Basic angle: use the positive value$\alpha = \cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)$$= 30^\circ$
   3. Step 3: ASTC (All, Sine, Tan, Cos): in which quadrants does cos have this sign?$\cos x$ is negative in quadrants 2 and 3.
   4. Step 4: List every angle x in the range, quadrant by quadrant$x = 180^\circ - 30^\circ$$= 150^\circ$$x = 180^\circ + 30^\circ$$= 210^\circ$