Lesson Solving trigonometric equations that include a multiple of an unknown as an argument

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<H2>Solving trigonometric equations that include a multiple of an unknown as an argument</H2>

<H3>Problem 1</H3>Solve trigonometric equation &nbsp;{{{(sin x + cos x)^2}}} = 1 &nbsp;for &nbsp;0 <= x < {{{2pi}}}.

<B>Solution</B>

<pre>
Your starting equation is

{{{(sin(x) + cos(x))^2}}} = 1

Transform it step by step

{{{sin^2(x) + 2*sin(x)*cos(x) + cos^2(x)}}} = 1,

1 + 2*sin(x)*cos(x) = 1,

sin(2x) = 0.

Hence, 2x should be a multiple of {{{pi}}}.

Since x should be in interval [{{{0}}},{{{2pi}}}),  2x should be in interval [{{{0}}},{{{4pi}}})

0 <= 2x < {{{4pi}}}.

There are 4 possible values for 2x: 0,  {{{pi}}},  {{{2pi}}}  and  {{{3pi}}}.

It gives 4 possible values for x:  0, {{{pi/2}}},  {{{pi}}}  and  {{{3pi/2}}}.

It is easy to check that all these 4 values satisfy given equation.

<U>ANSWER</U>.  Given equation has 4 solutions in the given interval:  0,  {{{pi/2}}}, {{{pi}}} and {{{3pi/2}}}.
</pre>

<H3>Problem 2</H3>Find all solutions to equation &nbsp;sec^2(x) + 2*tan(x) = 0 &nbsp;on interval &nbsp;[0, 2pi).

<B>Solution</B>

<pre>
Your starting equation is

sec^2 x + 2tan x = 0.

Transform it equivalently this way

{{{(1/cos^2(x))+(2sin(x)/cos(x))}}} = 0.

Now multiply both sides by cos(x).  Since cos(x) =/=0 by the meaning of the original equation,
we can do it secure in the equation's domain. We get then

1 + 2sin(x)*cos(x) = 0.

1 + sin(2x) = 0,

sin(2x) = -1

Hence,

2x = {{{3pi/2 + 2k*pi}}}, k = 0, +/-1, +/-2, . . .

We take only two values  2x = {{{3pi/2}}}  (k=0)  and  2x = {{{7pi/2}}}  (k=1),
since other values produce values of x out of the given interval.

<U>ANSWER</U>.  The given equation has two roots in the interval  [{{{0}}},{{{2pi}}}).

They are {{{3pi/4}}}, or 135°,  and  {{{7pi/4}}}, or 315°.
</pre>

<H3>Problem 3</H3>Find all solutions to equation &nbsp;&nbsp;{{{sqrt(3)*sec(4t)}}} = -2.

<B>Solution</B>

<pre>
Our starting equation is

sec(4t) = {{{-2/sqrt(3)}}}.

Transform it equivalently this way

cos(4t) = {{{-sqrt(3)/2}}}

4t = {{{5pi/6 + 2*pi*k}}}  OR  {{{7pi/6 + 2*pi*k}}},  'k'  is any integer.

Hence,

t = {{{5pi/24 + (pi/2)*k}}}  OR  t = {{{7pi/24+(pi/2)*k}}},  'k'  is any integer.    <U>ANSWER</U>
</pre>

<H3>Problem 4</H3>For equation &nbsp;cos(5x) = −1 &nbsp;find the smallest solution, the largest solution and the number of solutions for x in the interval &nbsp;0 ≤ x < 2π.

<B>Solution</B>

<pre>
The starting equation is

cos(5x) = -1.

The table solution for cosine is

5x = π, 3π, 5π, 7π, 9π, . . . (all odd integer multiples of π).

It implies

x = {{{pi/5}}}, {{{3pi/5}}},  {{{5pi/5}}} = π,   {{{7pi/5}}},  {{{9pi/5}}}, . . .

For greater or smaller values of 5x,  values of x will be out of the given interval - so, we do not consider them.

Thus, there are 5 solutions to the given equation in interval 0 <= x < 2π.

The smallest solution is  {{{pi/5}}}.  The greatest solution is  {{{9pi/5}}}.  The number of solutions is 5.  <U>ANSWER</U>
</pre>

<H3>Problem 5</H3>Find all solutions to equation &nbsp;&nbsp;{{{tan(theta/8)}}} + {{{sqrt3}}} = 0 &nbsp;in the interval &nbsp;[0, 2pi).

<B>Solution</B>

<pre>
tan(x/8) + √3 = 0,    (1)

tan(x/8) = -√3.       (2)

We are looking for possible solutions 'x' in interval [{{{0}}},{{{2pi}}}).

So, x/8 should be in interval [0,{{{(2pi)/8}}}), which is the same as interval [0,{{{pi/4}}}).

But in this interval from 0 to {{{pi/4}}}, function tan is always non-negative.

Hence, equation (2) with the negative right side HAS NO solutions in this interval.

It implies that the original equation has no solution/solutions in the given interval [0,2pi).     <U>ANSWER</U>
</pre>

<H3>Problem 6</H3>Find all exact solutions in interval &nbsp;0 ≤ x < 2π to equation
cos(2x)*cos(x) − sin(2x)*sin(x) = {{{sqrt(2)/ 2}}}.

<B>Solution</B>

<pre>
Use identity:  cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y)

cos(2x)* cos(x) − sin(2x)* sin( x) = square root of 2 over 2

cos(2x+x) = √2/2

2x+x = π/4 + 2k*π;     7π/4 + 2k*π.

3x = π/4 + 2k*π;     7π/4 + 2k*π.

For 3x, you must add the periods 2π and 4π for cosine.
    For 3x, it will produce geometrically the same angle; 
    but for 'x' it will produce new angles that you will miss otherwise.

You should no add more hire periods for 3x, since it will not produce 
    new angles for x and will lead you out of the given interval.

Now consider two cases.

Case (a).  3x = {{{pi/4}}},  {{{pi/4+2pi}}},  {{{pi/4+4pi}}}.

then  x = {{{pi/12}}},  {{{pi/12+(2pi)/3}}} = {{{9pi/12}}} = {{{(3/4)pi}}},  {{{pi/12+(4pi)/3}}} = {{{17pi/12}}}.

Case (b).  3x = {{{7pi/4}}},  {{{7pi/4+2pi}}},  {{{7pi/4+4pi}}}.

then  x = {{{7pi/12}}},  {{{7pi/12+(2pi)/3}}} = {{{15pi/12}}} = {{{(5/4)pi}}},  {{{7pi/12+(4pi)/3}}} = {{{23pi/12}}}.

<U>ANSWER</U>.  The given equation has 6 solutions in the given interval

{{{pi/12}}},  {{{(3/4)pi}}},  {{{17pi/12}}},  {{{7pi/12}}},  {{{(5/4)pi}}},  {{{23pi/12}}}.
</pre>

<H3>Problem 7</H3>cos(2x+20°) = sin(3x−10°). &nbsp;&nbsp;Find &nbsp;x.

<B>Solution</B>

<pre>
Equation

cos(2x+20°) = sin(3x−10°)    (1)

implies

(2x+20°) + (3x-10°) = = 90° + k*360°, k = 0, +/-1, +/-2, . . . (2)

because both functions tan and cot are periodical with period  {{{pi}}}.

So, we can rewrite equation (2) in the form

5x = 80° + k*360°, k = 0, 1, 2, 3, 4.

We can limit ourselves with 5 values of k, because for other values of 'k', the angles 'x' will be geometrically the same,
but formally will be outside of the interval [0°,360°).

For these 5 values of 'k', we will have 5 different values of x in the interval [0°,360°)

x = 16°,  16°+72° = 88°,  16°+144° = 160°,  16°+216° = 232°,  16°+288° = 304°.

<U>ANSWER</U>. Taking into account the periodicity of the left side and the right side of the given equation,

it is natural to look for solutions 'x' in the interval [0°,360°).

In this interval, the given equation has 5 (five) different solutions 16°, 88°, 160°, 232° and 302°.
</pre>

<H3>Problem 8</H3>tan(x) = cot(4x+20°).  &nbsp;&nbsp;Find &nbsp;x.

<B>Solution</B>

<pre>
Equation

tan(x) = cot(4x+20°)    (1)

implies

x + (4x+20°) = 90° + k*180°, k = 0, +/-1, +/-2, . . . (2)

because both functions tan and cot are periodical with period {{{pi}}}.

So, we can rewrite equation (2) in the form

5x = 70° + k*180°, k = 0, 1, 2, 3, 4.

We can limit ourselves with 5 values of k, because for other values of 'k', the angles 'x' will be geometrically the same,
but formally will be outside of the interval [0°,180°).

For these 5 values of 'k', we will have 5 different values of x in the interval [0°,180°)

x = 14°,  14°+36° = 50°,  14°+72° = 86°,  14°+108° = 122°,  14°+144° = 158°.

<U>ANSWER</U>. Taking into account the periodicity of the left side and the right side of the given equation,

it is natural to look for solutions 'x' in the interval [0°,180°).

In this interval, the given equation has 5 (five) different solutions 14°, 50°, 86°, 122° and 158°.
</pre>

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