Trigonometry

Higher Mathematics Unit 2 – Mathematics 24. Solve tan x° = − 5 for 0 < x < 360 .Notetan x° = − 5180° − x° x°̌ S AT C ̌180° + x° 360° − x°−1x = tan ( 5)= 78·690 (to 3 d.p.)x = 180 − 78·690 or 360 − 78·690x = 101·310 or 281·310Since tan x° is negativeAll trigonometric equations in Higher can be reduced to problems like thoseabove. The only differences are: the solutions could be required in radians – in this case, the question willnot have a degree symbol, e.g. Solve 3tan x = 1 rather than 3tan x° = 1 exact value solutions could be required in the non-calculator paperQuestions can be worked through in degrees or radians, but make sure thefinal answer is given in the units asked for in the question.EXAMPLES5. Solve 2sin2x° − 1 = 0 where 0 ≤ x ≤ 360 .2sin2x° = 1180° − 2x° 2x°̌ S A ̌ 0 ≤ x ≤ 360sin2x° = 12T C 0 ≤ 2x≤ 7202x = 30 or 180 − 30180° + 2x° 360° − 2x°2x= sin= 30or 360 + 30 or 360 + 180 − 30or 360 + 360 + 302x= 30 or 150 or 390 or 510x = 15 or 75 or 195 or 255−( )1 12NoteThere are more solutionsevery 360°, sincesin(30°) = sin(30°+360°) = …So keep adding 360 until2x > 720hsn.uk.netPage 91HSN22300