Trigonometric Equations I By Mr Porter

Basic Knowledge 1 1: Students need a basic understanding of solving a)Linear type equations: 3x + 1 = - 17 b)Quadratic Equations:2x 2 – 3x + 1 = 0 2: Students need a basic understanding of trigonometric identities:

Basic Knowledge 2 1: Students must know the exact values triangles used in trigonometry. 60° 30° 45° 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+)Tan(+) θ°180° — θ° 180° + θ° 360° — θ° 2: Students must know the acute angle equivalent of angles of any magnitude in trigonometry

Radian Angles Radian Angles are part of the HSC course, but some questions are written in radians. Radian Conversion π radians = 180° Degrees (θ) to Radians (x)Radian (x) to Degrees (θ) Note the following angles: Exact Values Triangles in Radians 0 or 2π Sin(+)All (+) Cos(+)Tan(+) xπ — x π + x2π – x

Example 1: Solve 2 sin θ – 1 = 0, for 0° ≤ θ ≤ 360° Solution: We can use the degree exact values triangles (or calculator) to find an angle that gives ½. 60° 30° 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+)Tan(+) θ°180° — θ° 180° + θ° 360° — θ° The trigonometric angle circle, tells us there are two possible POSITIVE sine angle ratios A and A So, the solutions are θ = 30°orθ = 150° Always check your answer(s) using a calculator! If you have a single trigonometric function, make it the subject of the equation. Degrees should be used in the solution.

Example 2: Solve 3 cos θ + 1 = 0, for 0° ≤ θ ≤ 360° Solution: 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+)Tan(+) θ°180° — θ° 180° + θ° 360° — θ° The trigonometric angle circle, tells us there are two possible NEGATIVE cosine angle ratios 180 – A and A So, the solutions are θ = 109°28’ orθ = 250°32’ Always check your answer(s) using a calculator! Note: 1/ 3 givens us the acute angle required. – tells us which Quadrant to look in. If you have a single trigonometric function, make it the subject of the equation. Degrees should be used in the solution.

Example 3: Solve 6 sin 2θ = 3, for Solution: Always check your answer(s) using a calculator! If you have a single trigonometric function, make it the subject of the equation. Radians should be used in the solution. The trigonometric angle circle, tells us there are two possible POSITIVE sine angle ratios A and π – A in 0 ≤ A ≤ 2π. Note: Original domain 0 ≤ θ ≤ 2π then for 2θ, domain is 0 ≤ 2θ ≤ 4π So, the solutions are Note: 1 / 2 givens us the acute angle required. + tells us which Quadrant to look in. 0 or 2π Sin(+)All (+) Cos(+)Tan(+) xπ — x π + x2π – x

Example 4: Solve, for Always check your answer(s) using a calculator! If you have a single trigonometric function, make it the subject of the equation. Degrees should be used in the solution. Note: Original domain 0° ≤ x ≤ 360° then for ˆx / 2, domain is 0° ≤ ˆx / 2 ≤ 180° Note: 3 / 4 givens us the acute angle required. + tells us which Quadrant to look in. Solution: The trigonometric angle circle, tells us there are two possible POSITIVE sine angle ratios A and 180° – A in 0° ≤ A ≤ 360°. 0°(360° ) 90° 270° 180° Sin(+)All (+) Cos(+)Tan(+) θ°180° — θ° 180° + θ° 360° — θ° So,A = 48°35 orA = 131°25’ So, subtracting 30°

Example 5: Solve, for If you have a single trigonometric function, make it the subject of the equation. Radians should be used in the solution. Note: Original domain 0 ≤ x ≤ 2π then for 2x, domain is 0 ≤ 2x ≤ 4π Solution: Solve for the equals case. Note: 1 / √2 givens us the acute angle required. + tells us which Quadrant to look in. 0 or 2π Sin(+)All (+) Cos(+)Tan(+) xπ — x π + x2π – x The trigonometric angle circle, tells us there are two possible POSITIVE cosine angle ratios A and 2π – A in 0 ≤ A ≤ 2π. To solve the inequality, you need to test in-between the above boundary points plus x = 0 and x = 2π OR use a neat sketch. Solutions are: