Station 1
Simplify the Expression
Sections 5.1
Q1
Standard
Simplify completely.
\[\sin x \csc x - \cos^2 x\]
Show your work
Q2
Intermediate
Simplify completely. Write your answer without fractions.
\[\dfrac{\sin^2 x}{1 + \cos x}\]
Show your work
Q3
Challenge
Simplify completely.
\[\dfrac{1 + \tan^2 x}{1 + \cot^2 x}\]
Show your work
Station 2
Verify the Identity
Section 5.2
Q1
Standard
Verify the identity. Show all steps working one side only.
\[\sin x \sec x = \tan x\]
Show your work
Q2
Intermediate
Verify the identity. Show all steps working one side only.
\[\dfrac{\cos x}{1 - \sin x} = \sec x + \tan x\]
Show your work
Q3
Challenge
Verify the identity. Show all steps working one side only.
\[(\csc x - \cot x)^2 = \dfrac{1 - \cos x}{1 + \cos x}\]
Show your work
Station 3
Solve on \([0, 2\pi)\)
Section 5.3
Q1
Standard
Find all solutions in \([0, 2\pi)\).
\[\tan^2 x - 1 = 0\]
Show your work
Q2
Intermediate
Find all solutions in \([0, 2\pi)\).
\[2\cos^2 x + 3\cos x + 1 = 0\]
Show your work
Q3
Challenge
Find all solutions in \([0, 2\pi)\).
\[\cos^2 x - 3\sin x + 3 = 0\]
Show your work
Station 4
Solve — General Solution
Section 5.3
Q1
Standard
Find all solutions. Write your answer in general form.
\[2\cos x + \sqrt{3} = 0\]
Show your work
Q2
Intermediate
Find all solutions. Write your answer in general form.
\[\sin x \tan x = \sin x\]
Show your work
Q3
Challenge
Find all solutions. Write your answer in general form.
\[2\cos^2 x - 3\sin x = 3\]
Show your work
Answer Key — Stations 1 & 2
Station 1 — Simplify
Q1
\(\sin x \csc x - \cos^2 x = 1 - \cos^2 x = \boxed{\sin^2 x}\)
Q2
\(\dfrac{\sin^2 x}{1 + \cos x} = \dfrac{(1-\cos x)(1+\cos x)}{1+\cos x} = \boxed{1 - \cos x}\)
Q3
\(\dfrac{1 + \tan^2 x}{1 + \cot^2 x} = \dfrac{\sec^2 x}{\csc^2 x} = \dfrac{\sin^2 x}{\cos^2 x} = \boxed{\tan^2 x}\)
Station 2 — Verify
Q1
LHS: \(\sin x \cdot \dfrac{1}{\cos x} = \dfrac{\sin x}{\cos x} = \tan x\;\checkmark\)
Q2
Work the left side — multiply by conjugate \(\dfrac{1+\sin x}{1+\sin x}\):
\(\dfrac{\cos x(1+\sin x)}{1-\sin^2 x} = \dfrac{\cos x(1+\sin x)}{\cos^2 x} = \dfrac{1+\sin x}{\cos x} = \sec x + \tan x\;\checkmark\)
Q3
Work the left side — convert to sin & cos, then square:
\(\left(\dfrac{1-\cos x}{\sin x}\right)^2 = \dfrac{(1-\cos x)^2}{\sin^2 x} = \dfrac{(1-\cos x)^2}{1-\cos^2 x} = \dfrac{(1-\cos x)^2}{(1-\cos x)(1+\cos x)} = \dfrac{1-\cos x}{1+\cos x}\;\checkmark\)
Answer Key — Stations 3 & 4
Station 3 — Solve on \([0, 2\pi)\)
Q1
\(\tan^2 x = 1 \;\Rightarrow\; \tan x = \pm 1 \;\Rightarrow\; \boxed{x = \dfrac{\pi}{4},\; \dfrac{3\pi}{4},\; \dfrac{5\pi}{4},\; \dfrac{7\pi}{4}}\)
Q2
Factor: \((2\cos x + 1)(\cos x + 1) = 0\)
\(\cos x = -\tfrac{1}{2} \;\Rightarrow\; x = \tfrac{2\pi}{3},\, \tfrac{4\pi}{3}\)    \(\cos x = -1 \;\Rightarrow\; x = \pi\)    \(\boxed{x = \dfrac{2\pi}{3},\; \pi,\; \dfrac{4\pi}{3}}\)
Q3
Pythagorean sub: \((1 - \sin^2 x) - 3\sin x + 3 = 0\)
\(\sin^2 x + 3\sin x - 4 = 0 \;\Rightarrow\; (\sin x + 4)(\sin x - 1) = 0\)
\(\sin x = -4\) (impossible)    \(\sin x = 1 \;\Rightarrow\;\) \(\boxed{x = \dfrac{\pi}{2}}\)
Station 4 — General Solution
Q1
\(\cos x = -\dfrac{\sqrt{3}}{2} \;\Rightarrow\; \boxed{x = \dfrac{5\pi}{6} + 2\pi n \;\;\text{or}\;\; x = \dfrac{7\pi}{6} + 2\pi n}\)
Q2
Factor (don't divide!): \(\sin x(\tan x - 1) = 0\)
\(\sin x = 0 \;\Rightarrow\; x = \pi n\)    \(\tan x = 1 \;\Rightarrow\; x = \dfrac{\pi}{4} + \pi n\)    \(\boxed{x = \pi n \;\;\text{or}\;\; x = \dfrac{\pi}{4} + \pi n}\)
Q3
Pythagorean sub: \(2(1-\sin^2 x) - 3\sin x - 3 = 0 \;\Rightarrow\; 2\sin^2 x + 3\sin x + 1 = 0\)
\((2\sin x + 1)(\sin x + 1) = 0\)
\(\sin x = -\tfrac{1}{2} \;\Rightarrow\; x = \tfrac{7\pi}{6} + 2\pi n\) or \(x = \tfrac{11\pi}{6} + 2\pi n\)
\(\sin x = -1 \;\Rightarrow\; x = \tfrac{3\pi}{2} + 2\pi n\)
\(\boxed{x = \dfrac{7\pi}{6} + 2\pi n, \;\; \dfrac{3\pi}{2} + 2\pi n, \;\; \dfrac{11\pi}{6} + 2\pi n}\)