Solving trigonometric equations is a common topic on the examination. The key to solving them is a good knowledge of the trigonometric functions. Whether you prefer to use the Unit Circle or the graphs of the functions, you need to a method that works for you. This topic often comes up on the non-calculator paper. Therefore, it is important that you do not solely rely on your graphical calculator to solve the equations and you need to remember the exact values for sin/cos/tan of 30°, 45°, 60°, ...
On this page, you should learn about
- solving trigonometric equations (including quadratic) both graphically and analytically
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Here is a quiz that practises the skills from this page
START QUIZ!
Some of the following statements are TRUE and some are FALSE.
Type T (TRUE) or F (FALSE) next to each one
a. sin x = sin (180 - x)
b. cos x = cos (180 - x)
c. sin x = sin (180 + x)
d. cos x = cos (360 + x)
e. tan x = tan (180 + x)
f. cos (180 - x) = cos (180 + x)
g. sin x = sin (-x)
h. sin x = cos (x - 180)
i. cos x = cos ( -x)
j. tan (180 - x) = tan (180 + x)
Some of the following statements are TRUE and some are FALSE.
Type T (TRUE) or F (FALSE) next to each one
a. sin (-x) = sin (\(2\pi\) - x)
b. sin (\(\pi\) - x) = sin (\(2\pi\) + x)
c. tan x = tan (\(\pi\) - x)
d. tan x = tan (-x)
e. cos (-x) = -cosx
f. tan x = tan (\(2\pi\) + x)
g. cos x = cos (\(2\pi\) - x)
h. cos x = cos (\(\pi\) + x)
i. tan (\(\pi\) - x) = tan (\(\pi\) + x)
j. cos (\(2\pi\) + x) = cos (\(2\pi\) - x))
Complete all the equations below using a different angle.
All angles are between 0° and 360°
sin 40° = sin °
cos 80° = cos °
tan 25° = tan °
sin 100° = sin °
tan 269° = tan °
sinx = sin(180 - x)
cosx = cos(360 - x)
tanx = tan(180 + x)
Complete all the equations below using a different angle.
All angles are between 0° and 360°
cos 359° = cos °
cos 135° = cos °
tan 102° = tan °
sin 208° = sin °
cos 251° = cos °
sinx = sin(180 - x)
cosx = cos(360 - x)
tanx = tan(180 + x)
Complete the equations below using a different angle. All the angles are between 0 and \(2\pi\)
Give your answers correct to 3 significant figures.
cos 1.15 = cos
sin 0.834 = sin
tan 0.335 = tan
tan 4.26 = tan
cos 5.71 = cos
sinx = sin(\(\pi\) - x)
cosx = cos(\(2\pi\) - x)
tanx = tan(\(\pi\) + x)
Complete the equations below using a different angle. All the angles are between 0 and \(2\pi\)
Give your answers correct to 3 significant figures.
sin 3.40 = sin
tan 2.99 = tan
cos 2.02 = cos
sin (\(\pi\) + 2.58) = sin(2\(\pi\) - 2.58)
tan(\(\pi\) - 0.152) = tan(\(2\pi\) - 0.152)
cos(\(\pi\) - 1.12) = cos(\(\pi\) + 1.12)
Solve \(sin\theta = \frac{\sqrt{2}}{2}\) for 0° < \(\theta\) < 360°
\(\theta_1<\theta_2\)
\(\theta_1\) = °
\(\theta_2\) = °
\(\theta_1\) = 45°
\(\theta_2\)= 180° - 45°
The solutions to \(cos\theta = -\frac{\sqrt{3}}{2}\) for 0 < \(\theta\) < \(2\pi\)
are \(\frac { a\pi }{ 6 } \) and \(\frac { b\pi }{ 6 } \) , where a < b
Find a and b
a =
b =
\(arccos(+ \frac{\sqrt{3}}{2})=\frac { \pi }{ 6} \)
\(\theta_1\) = \(\pi-\frac { \pi }{ 6 } \) = \(\frac { 5\pi }{ 6 } \)
\(\theta_2\)= \(\pi+\frac { \pi }{ 6 } \) = \(\frac { 7\pi }{ 6 } \)
The solutions to \(sin2\theta = -\frac{1}{2}\) for 0° < \(\theta\) < \(2\pi\)
are \(\frac { a\pi }{ 12 } ,\frac { b\pi }{ 12 } ,\frac { c\pi }{ 12 } ,\frac { d\pi }{ 12 } \) , where a < b < c < d
Find a , b , c and d
a =
b =
c =
d =
\(sin2\theta = -\frac{1}{2}\) for 0° < \(2\theta\) < \(4\pi\)
\(arcsin(+ \frac{1}{2})=\frac{\pi}{6}\)
\(2\theta=\frac { \pi }{ 6 }\quad ,\quad \pi-\frac { \pi }{ 6 }\quad ,\quad 2\pi+\frac { \pi }{ 6 }\quad ,\quad 2\pi+(\pi-\frac { \pi }{ 6 })\)
\(2\theta =\frac { \pi }{ 6 } ,\frac { 5\pi }{ 6 } ,\frac { 13\pi }{ 6 } ,\frac { 17\pi }{ 6 } \)
\(\theta =\frac { \pi }{12 } ,\frac { 5\pi }{ 12 } ,\frac { 13\pi }{ 12 } ,\frac { 17\pi }{ 12 } \)
Solve \(tan^2\theta = 1\) for -180° < \(\theta\) < 180°
\(\theta_1<\theta_2<\theta_3<\theta_4\)
\(\theta_1\)= °
\(\theta_2\) = °
\(\theta_3\) = °
\(\theta_4\) = °
\(tan^2\theta = 1\)
\(tan\theta =\pm 1\)
arctan(+1) = 45°
Let f(x)= cosx and g(x) = \(\frac{2x^2}{1-x}\)
a) Show that g∘f(x) = 1 can be written as 2cos²x + cosx - 1 = 0
b) Hence solve g∘f(x)=1 for \(-\pi\le x\le \pi\)
Hint
Full Solution
Solve \(\log _{ 3 }{ sinx-\log _{ 3 }{ cosx=0.5 } } \) for \(0\le x\le 2\pi\)
Hint
Full Solution
How much of Solving Trigonometric Equations have you understood?
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