Examples – Trigonometric equations
Based on what we have explained to the article Trigonometric equations, we are going to solve some exercises below:
Example 1: Solve the equations.
a) $ \displaystyle \sin x=\sin {{10}^{{}^\circ }}$
Since the angles $ \displaystyle x$ and $ \displaystyle {{10}^{{}^\circ }}$ have the same sinuses, they are connected with the equations:
$ \displaystyle x=k\cdot {{360}^{\circ }}+{{10}^{{}^\circ }}$ and $ \displaystyle x=k\cdot {{360}^{\circ }}+({{180}^{{}^\circ }}-{{10}^{{}^\circ }})$ $ \displaystyle k\in Z$
All the solutions of this equation are:
$ \displaystyle x=k\cdot {{360}^{\circ }}+{{10}^{{}^\circ }}$ and $ \displaystyle x=k\cdot {{360}^{\circ }}+{{170}^{{}^\circ }}$
b) $ \displaystyle \sin 3x=\sin x$
Since the angles $ \displaystyle3x$ and $ \displaystyle x$ have the same sinuses, they are connected with the equations:
$ \displaystyle 3x=k\cdot {{360}^{\circ }}+x$ and $ \displaystyle x=k\cdot {{360}^{\circ }}+({{180}^{{}^\circ }}-x)$
All the solutions of this equation are:
$ \displaystyle x=k\cdot {{180}^{{}^\circ }}$ and $ \displaystyle x=k\cdot {{90}^{{}^\circ }}+{{45}^{{}^\circ }}$
Example 2: Solve the equations for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{360}^{\circ }}$.
a) $ \displaystyle \sin (x+{{30}^{\circ }})=\frac{1}{2}$
An angle that has the sinuses $ \displaystyle \frac{1}{2}$ is $ \displaystyle {{30}^{\circ }}$
Another angle is $ \displaystyle {{180}^{\circ }}-{{30}^{\circ }}={{150}^{{}^\circ }}$
The general solutions are:
$ \displaystyle x+{{30}^{\circ }}=k\cdot {{360}^{\circ }}+{{30}^{\circ }}$
$ \displaystyle x=k\cdot {{360}^{\circ }}$ and $ \displaystyle x+{{30}^{\circ }}=k\cdot {{360}^{\circ }}+{{150}^{\circ }}$
$ \displaystyle x=k\cdot {{360}^{\circ }}+{{120}^{{}^\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{360}^{\circ }}$ are:
For $ \displaystyle k=0$ and $ \displaystyle k=1$, $ \displaystyle \left\{ {{{0}^{{}^\circ }}{{{,120}}^{{}^\circ }}{{{,360}}^{\circ }}} \right\}$
b) $ \displaystyle \cos (x-{{30}^{\circ }})=\frac{1}{2}$
An angle that has the cosines $ \displaystyle \frac{1}{2}$ is $ \displaystyle {{60}^{\circ }}$
Another angle is $ \displaystyle-{{60}^{\circ }}$
The general solutions are:
$ \displaystyle x-{{30}^{\circ }}=k\cdot {{360}^{\circ }}+{{60}^{\circ }}$
$ \displaystyle x=k\cdot {{360}^{\circ }}+{{90}^{\circ }}$ and $ \displaystyle x-{{30}^{\circ }}=k\cdot {{360}^{\circ }}-{{60}^{\circ }}$
$ \displaystyle x=k\cdot {{360}^{\circ }}-{{30}^{\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{360}^{\circ }}$ are:
For $ \displaystyle k=0$ and $ \displaystyle k=1$, $ \displaystyle \left\{ {{{{90}}^{\circ }}{{{,330}}^{{}^\circ }}} \right\}$
c) $ \displaystyle tg(x+{{20}^{{}^\circ }})=1$
An angle that has the tangent $ \displaystyle 1$ is $ \displaystyle {{45}^{{}^\circ }}$
The general solutions are:
$ \displaystyle x+{{20}^{{}^\circ }}=k\cdot {{180}^{{}^\circ }}+{{45}^{{}^\circ }}$
$ \displaystyle x=k\cdot {{180}^{{}^\circ }}+{{25}^{{}^\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{360}^{\circ }}$ are:
For $ \displaystyle k=0$ and $ \displaystyle k=1$, $ \displaystyle \left\{ {{{{25}}^{\circ }}{{{,205}}^{{}^\circ }}} \right\}$
d) $ \displaystyle 2\sin (x+{{30}^{\circ }})=-1$
$ \displaystyle \sin (x+{{30}^{\circ }})=-\frac{1}{2}$
An angle that has the sinuses$ \displaystyle -\frac{1}{2}$ is $ \displaystyle -{{30}^{{}^\circ }}$
Another angle is $ \displaystyle {{180}^{\circ }}-(-{{30}^{{}^\circ }})={{210}^{{}^\circ }}$
The general solutions are:
$ \displaystyle x+{{30}^{\circ }}=k\cdot {{360}^{\circ }}-{{30}^{\circ }}$
$ \displaystyle x=k\cdot {{360}^{\circ }}-{{60}^{\circ }}$ and $ \displaystyle x+{{30}^{\circ }}=k\cdot {{360}^{\circ }}+{{210}^{\circ }}$
$ \displaystyle x=k\cdot {{360}^{\circ }}+{{180}^{\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{360}^{\circ }}$ are:
For $ \displaystyle k=0$ and $ \displaystyle k=1$, $ \displaystyle \left\{ {{{{180}}^{{}^\circ }}~or~~{{{300}}^{{}^\circ }}} \right\}$
Example 3: Solve the equations for$ \displaystyle {{0}^{{}^\circ }}\le x\le {{180}^{\circ }}$.
a) $ \displaystyle 2\sin 2x=1$
$ \displaystyle \sin 2x=\frac{1}{2}$
An angle that has the sinuses $ \displaystyle \frac{1}{2}$ is the angle $ \displaystyle {{30}^{{}^\circ }}$
Another angle is $ \displaystyle {{180}^{{}^\circ }}-{{30}^{{}^\circ }}={{150}^{{}^\circ }}$
The general solutions are:
$ \displaystyle 2x=k\cdot {{360}^{\circ }}+{{30}^{{}^\circ }}$
$ \displaystyle x=k\cdot {{180}^{{}^\circ }}+{{15}^{{}^\circ }}$ and $ \displaystyle 2x=k\cdot {{360}^{\circ }}+{{150}^{{}^\circ }}$
$ \displaystyle x=k\cdot {{180}^{{}^\circ }}+{{75}^{{}^\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{180}^{\circ }}$.
For $ \displaystyle k=0$, $ \displaystyle \left\{ {{{{15}}^{{^{{}^\circ }}}},{{{75}}^{{^{{}^\circ }}}}} \right\}$
b) $ \displaystyle 3tg2x=2$
$ \displaystyle tg2x=\frac{2}{3}$
An angle that has the tangent $ \displaystyle \frac{2}{3}$ is $ \displaystyle {{33,5}^{{}^\circ }}$
The general solutions are:
$ \displaystyle 2x=k\cdot {{180}^{{}^\circ }}+{{33,5}^{{}^\circ }}$
$ \displaystyle x=k\cdot {{90}^{\circ }}+{{16.75}^{{}^\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{180}^{\circ }}$.
For $\displaystyle k=0$, $\displaystyle \left\{ {{{{16.75}}^{{^{{}^\circ }}}}~~or~~{{{106.75}}^{{^{{}^\circ }}}}} \right\}$
c) $ \displaystyle 5\cos 3x=2$
$ \displaystyle \cos 3x=\frac{2}{5}$
An angle that has the cosine is the angle$ \displaystyle {{66}^{\circ }}$
Another angle is $ \displaystyle-{{66}^{\circ }}$
The general solutions are:
$ \displaystyle 3x=k\cdot {{360}^{\circ }}+{{66}^{\circ }}$
$ \displaystyle x=k\cdot {{120}^{{}^\circ }}+{{33}^{{}^\circ }}$ and $ \displaystyle 3x=k\cdot {{360}^{\circ }}-{{66}^{\circ }}$
$ \displaystyle x=k\cdot {{120}^{{}^\circ }}-{{33}^{{}^\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{180}^{\circ }}$
For $ \displaystyle k=0$ and $ \displaystyle k=1$, $ \displaystyle \left\{ {{{{33}}^{{}^\circ }}{{{,87}}^{{}^\circ }}{{{,153}}^{{}^\circ }}} \right\}$
d) $ \displaystyle 5\sin 3x+3=0$
$ \displaystyle 5\sin 3x=-3$
$ \displaystyle \sin 3x=-\frac{3}{5}$
An angle that has the sinuses $ \displaystyle -\frac{3}{5}$ is $ \displaystyle -{{37}^{{}^\circ }}$
Another angle is the angle $ \displaystyle {{180}^{{}^\circ }}-(-{{37}^{{}^\circ }})={{217}^{{}^\circ }}$
The general solutions are:
$ \displaystyle 3x=k\cdot {{360}^{{}^\circ }}-{{37}^{{}^\circ }}$
$ \displaystyle x=k\cdot {{120}^{{}^\circ }}-{{12.3}^{{}^\circ }}$ and $ \displaystyle 3x=k\cdot {{360}^{{}^\circ }}+{{217}^{{}^\circ }}$
$ \displaystyle x=k\cdot {{120}^{{}^\circ }}+{{72.3}^{{}^\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{180}^{\circ }}$.
For $ \displaystyle k=0$ and $ \displaystyle k=1$, $ \displaystyle \left\{ {{{{72.3}}^{{}^\circ }}{{{,107.7}}^{{}^\circ }}} \right\}$
Example 4: Solve the equations for $ \displaystyle -{{180}^{{}^\circ }}\le x\le {{180}^{{}^\circ }}$.
a) $ \displaystyle 2\sin x=\cos x$
Firstly we transform our equation by dividing with $ \displaystyle \cos x$ both sides.
We should note that $ \displaystyle \cos x\ne 0$
$ \displaystyle x\ne k\cdot {{360}^{{}^\circ }}\pm {{90}^{{}^\circ }}$
But for $ \displaystyle -{{180}^{{}^\circ }}\le x\le {{180}^{{}^\circ }}$ then $ \displaystyle x\ne \pm {{90}^{{}^\circ }}$
$ \displaystyle 2\sin x=\cos x$
$ \displaystyle \frac{{2\sin x}}{{\cos x}}=\frac{{\cos x}}{{\cos x}}$
$ \displaystyle 2tgx=1$
$ \displaystyle tgx=\frac{1}{2}$
An angle that has the tangent approximately $ \displaystyle \frac{1}{2}$ is the angle $ \displaystyle {{27}^{{}^\circ }}$
The general solutions are:
$ \displaystyle x=k\cdot {{180}^{{}^\circ }}+{{27}^{{}^\circ }}$
The solutions for $ \displaystyle -{{180}^{{}^\circ }}\le x\le {{180}^{{}^\circ }}$
For $ \displaystyle k=0$ and $ \displaystyle k=-1$, $ \displaystyle \left\{ {{{{27}}^{{}^\circ }}{{{,153}}^{{}^\circ }}} \right\}$
b) $ \displaystyle 3\sin 2x=2tg2x$
Firstly we transform our equation from both sides:
$ \displaystyle 3\sin 2x=2tg2x$
$ \displaystyle 3\sin 2x=2\frac{{\sin 2x}}{{\cos 2x}}$
$ \displaystyle 3\sin 2x\cdot \cos 2x=2\sin 2x$
We divide with $ \displaystyle {2\sin 2x}$ and note that:
$\displaystyle 2\sin 2x\ne 0$
$\displaystyle \sin 2x\ne 0$
$\displaystyle 2x\ne k\cdot {{360}^{{}^\circ }}$
$\displaystyle x\ne k\cdot {{180}^{{}^\circ }}$ and $\displaystyle 2x\ne k\cdot {{360}^{{}^\circ }}+{{180}^{{}^\circ }}$
$\displaystyle x\ne k\cdot {{180}^{{}^\circ }}+{{90}^{\circ }}$
$ \displaystyle \frac{{3\sin 2x\cdot \cos 2x}}{{2\sin 2x}}=\frac{{2\sin 2x}}{{2\sin 2x}}$
$ \displaystyle \frac{{3\cos 2x}}{2}=1$
$ \displaystyle \cos 2x=\frac{2}{3}$
An angle that has the cosines approximately $ \displaystyle \frac{2}{3}$ is $ \displaystyle {{48}^{{}^\circ }}$.
Another angle is $ \displaystyle-{{48}^{{}^\circ }}$
The general solutions are:
$ \displaystyle 2x=k\cdot {{360}^{{}^\circ }}\pm {{48}^{{}^\circ }}$
$ \displaystyle x=k\cdot {{120}^{{}^\circ }}\pm {{24}^{{}^\circ }}$
The solutions for $ \displaystyle -{{180}^{{}^\circ }}\le x\le {{180}^{{}^\circ }}$
For $ \displaystyle k=0$ ,$ \displaystyle k=-1$ and $ \displaystyle k=1$ are:
$ \displaystyle \left\{ {\pm {{{24}}^{{}^\circ }},\pm {{{144}}^{{}^\circ }}} \right\}$
c) $ \displaystyle 3\sin x+tgx=0$
Firstly we transform our equation from both sides:
$ \displaystyle 3\sin x+tgx=0$
$ \displaystyle 3\sin x=-tgx$
$\displaystyle 3\sin x=-\frac{{\sin x}}{{\cos x}}$
We divide with $\displaystyle {\sin x}$ and note that $\displaystyle \sin x\ne 0$
$\displaystyle x\ne k\cdot {{360}^{\circ }}$ and $\displaystyle x\ne k\cdot {{360}^{\circ }}+{{180}^{\circ }}$
$\displaystyle \frac{{3\sin x\cos x}}{{\sin x}}=\frac{{-\sin x}}{{\sin x}}$
$\displaystyle 3\cos x=-1$
$\displaystyle \cos x=-\frac{1}{3}$
An angle that has the cosine $\displaystyle -\frac{1}{3}$ is approximately $\displaystyle {{109}^{{}^\circ }}$
Another angle is the angle $\displaystyle-{{109}^{{}^\circ }}$
The general solutions are:
$\displaystyle x=k\cdot {{360}^{{}^\circ }}\pm {{109}^{{}^\circ }}$
The solutions for $ \displaystyle -{{180}^{{}^\circ }}\le x\le {{180}^{{}^\circ }}$
For $ \displaystyle k=0$ the solutions are: $\displaystyle \left\{ {{{{109}}^{{}^\circ }},-{{{109}}^{{}^\circ }}} \right\}$
d) $ \displaystyle \sin x\cos x-\cos x=0$
Firstly we transform our equation from both sides:
$\displaystyle \sin x\cos x=\cos x$
Then we divide with $\displaystyle \cos x$ and note that $\displaystyle \cos x\ne 0$
$ \displaystyle x\ne k\cdot {{360}^{\circ }}\pm {{90}^{\circ }}$
$\displaystyle \frac{{\sin x\cos x}}{{\cos x}}=\frac{{\cos x}}{{\cos x}}$
$\displaystyle \sin x=1$
The general solutions are:
$\displaystyle x=k\cdot {{360}^{\circ }}+{{90}^{\circ }}$
The solutions for $ \displaystyle -{{180}^{{}^\circ }}\le x\le {{180}^{{}^\circ }}$
For $ \displaystyle k=0$the solutions are: $ \displaystyle \left\{ {{{{90}}^{\circ }}} \right\}$
Example 5: Solve the equations for $ \displaystyle {{0}^{{}^\circ }}\le x\le {{360}^{{}^\circ }}$.
a) $ \displaystyle {{\sin }^{2}}x-2\sin x+1=0$
As we see we have a quadratic trigonometric equation, and to make it easy for solving we substitute $\displaystyle \sin x=t$ and solve our equation firstly in relation with t.
$ \displaystyle {{\sin }^{2}}x-2\sin x+1=0$
$\displaystyle {{t}^{2}}-2t+1=0$
To see how to solve it Check Quadratic equations
$ \displaystyle t=1$ so $ \displaystyle \sin x=1$ and all the solutions for $ \displaystyle {{0}^{{}^\circ}}\le x\le {{360}^{{}^\circ }}$ are: $ \displaystyle x=k\cdot {{360}^{\circ }}+{{90}^{\circ }}$
b) $ \displaystyle 2\sin x+2=3{{\cos }^{2}}x$
Firstly we substitute $ \displaystyle {{\cos }^{2}}x=1-{{\sin }^{2}}x$ to turn it into a quadratic equation of the same trigonometric function.
$ \displaystyle 2\sin x+2=3{{\cos }^{2}}x$
$ \displaystyle 2\sin x+2=3(1-{{\sin }^{2}}x)$
$ \displaystyle 2\sin x+2=3-3{{\sin }^{2}}x$
$ \displaystyle 3{{\sin }^{2}}x+2\sin x-1=0$
Then we substitute $ \displaystyle \sin x=t$ and solve the quadratic equation:
$ \displaystyle 3{{t}^{2}}+2t-1=0$
To see how to solve it Check Quadratic equations
$ \displaystyle {{t}_{1}}=-1$ so $ \displaystyle \sin x=-1$ the general solutions are approximately: $ \displaystyle x=k\cdot {{360}^{\circ }}+{{270}^{\circ }}$
$ \displaystyle {{t}_{2}}=\frac{1}{3}$ so $ \displaystyle \sin x=\frac{1}{3}$ the general solutions are approximately: $ \displaystyle x=k\cdot {{360}^{\circ }}-{{19}^{\circ }}$ and $ \displaystyle x=k\cdot {{360}^{\circ }}-{{161}^{\circ }}$d
The solutions for $ \displaystyle {{0}^{{}^\circ}}\le x\le {{360}^{{}^\circ }}$ are approximately: $ \displaystyle \left\{ {{{{19}}^{{}^\circ }}{{{,161}}^{{}^\circ }}{{{,270}}^{{}^\circ }}} \right\}$
c) $ \displaystyle 2{{\cos }^{2}}x+{{\sin }^{2}}x=2$
Firstly we substitute one of the functions to turn it into a quadratic equation of the same function:
$ \displaystyle {{\sin }^{2}}x=1-{{\cos }^{2}}x$
$ \displaystyle 2{{\cos }^{2}}x+1-{{\cos }^{2}}x=2$
$ \displaystyle {{\cos }^{2}}x=1$
$ \displaystyle \cos x=\pm 1$
The general solutions are:
For $ \displaystyle \cos x=1$ then $ \displaystyle x=k\cdot {{360}^{{}^\circ }}$ and
$ \displaystyle \cos x=-1$ then $ \displaystyle x=k\cdot {{360}^{{}^\circ }}+{{180}^{\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ}}\le x\le {{360}^{{}^\circ }}$ are:
$ \displaystyle \left\{ {{{0}^{{}^\circ }}{{{,180}}^{\circ }}{{{,360}}^{{}^\circ }}} \right\}$
d) $ \displaystyle t{{g}^{2}}x-tgx-2=0$
Firstly we substitute $ \displaystyle tgx=t$ and rewrite our quadratic equation:
$ \displaystyle {{t}^{2}}-t-2=0$
To see how to solve it Check Quadratic equations
$ \displaystyle {{t}_{1}}=2$ so $ \displaystyle tgx=2$ and the general solution are approximately:
$ \displaystyle x=k\cdot {{180}^{\circ }}+{{64}^{{}^\circ }}$
$ \displaystyle {{t}_{2}}=-1$ so $ \displaystyle tgx=-1$ and the general solutions are approximatelly: $ \displaystyle x=k\cdot {{180}^{\circ }}+{{135}^{{}^\circ }}$
The solutions for $ \displaystyle {{0}^{{}^\circ}}\le x\le {{360}^{{}^\circ }}$ are: $ \displaystyle \left\{ {{{{64}}^{{}^\circ }}{{{,135}}^{{}^\circ }}{{{,244}}^{{}^\circ }}{{{,315}}^{{}^\circ }}} \right\}$