##### 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: $\displaystyle \left\{ {{{{64}}^{{}^\circ }}{{{,135}}^{{}^\circ }}{{{,244}}^{{}^\circ }}{{{,315}}^{{}^\circ }}} \right\}$