Trigonometric Equations with Multiple Angles The diagram shows

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Trigonometric Equations with Multiple Angles

Trigonometric Equations with Multiple Angles

The diagram shows where the various ratios are positive Note also that values repeat

The diagram shows where the various ratios are positive Note also that values repeat every 360 o 90° 90 180 - S A 180° 180 + 0°, 360° 360 C T 270° 270 360 -

Solve cos 3 x = – 1/√ 2 for values of x such that

Solve cos 3 x = – 1/√ 2 for values of x such that 0 < x < π First of all consider the range of values If 0 < x < π then 0 < 3 x < 3π (540 o) o = to o Work in Remember degrees then 3π change = 3 × 180 back 540 radians later Now find BASE angle cos -1 (1/√ 2 ) = 45 o π/ 4 S 180 – A Since cosine negative T C o o o 3 x = 135 , 225 495 180 + Now change back to radians using multiples of π/4 3 x = 3π/4 , 5π/4 , 11π/4 So x = 3π/12 , 5π/12 , 11π/12

Solve sin 2 x = 1/2 for values of x such that 0 <

Solve sin 2 x = 1/2 for values of x such that 0 < x < 2π First of all consider the range of values If 0 < x < 2π then 0 < 2 x < 4π (720 o) o = to o Work in Remember degrees. Now then find change back angle radians later 4π = 4 BASE × 180 720 sin -1 (1/2 ) = 30 o π/ 6 S 180 – A Since sine positive T C o o 2 x = 30 , 150 390 , 510 Now change back to radians using multiples of π/6 2 x = π/6 , 5π/6 , 13π/6 , 17π/6 So x = π/12 , 5π/12 , 13π/12 , 17π/12

Solve tan 4 x = √ 3 for values of x such that 0

Solve tan 4 x = √ 3 for values of x such that 0 < x < π First of all consider the range of values If 0 < x < π then 0 < 4 x < 4π (720 o) o =to o Work in degrees Remember Now then find 4πchange =BASE 4 × 180 back angle 720 radians later tan -1 (√ 3 ) = 60 o π/ 3 S A Since tangent positive T C 180 + o o 4 x = 60 , 240 420 , 600 Now change back to radians using multiples of π/3 4 x = π/3 , 4π/3 , 7π/3 , 10π/3 So x = π/12 , π/3 , 7π/12 , 5π/6

Solve 2 sin (2 x – π/6) = √ 3 for values of x

Solve 2 sin (2 x – π/6) = √ 3 for values of x such that 0 < x < 2π First of all consider the range of values If 0 < x < 2π then 0 < 2 x – π/6 < 4π – π/6) (690 o) π/ change o – to o = 690 olater Work Remember in degrees Now 2πthen –find = BASE 2 × 180 back angle 30 radians 6 sin -1 (√ 3/2 ) = 60 o 2 x – π/ 6 = 60 o , π/ 120 o 180 - 3 420 o , 480 o S A T C Now change back to radians using multiples of π/3 2 x – π/6 = π/3 , 2π/3 , 7π/3 , 8π/3 2 x = 3π/ 6 , 5π/ 6 , 15π/ 6 , 17π/ 6 So x = π/4 , 5π/12 , 5π/6 , 17π/12