Explore Solving Trigonometric Equations with Inverse Functions and Identities

Here are a few recommended readings before getting started with this lesson.

Trigonometric Function	Inverse Trigonometric Function
$f (x) = sin x$	$f^{- 1} (x) = sin^{- 1} x$
$f (x) = cos x$	$f^{- 1} (x) = cos^{- 1} x$
$f (x) = tan x$	$f^{- 1} (x) = tan^{- 1} x$

Inverse Trigonometric Function	Domain	Range
$y = sin^{- 1} x$	$[- 1, 1]$	$- \frac{π}{2} \leq x \leq \frac{π}{2}$
$y = cos^{- 1} x$	$[- 1, 1]$	$0 \leq x \leq π$
$y = tan^{- 1} x$	All real numbers	$- \frac{π}{2} \leq x \leq \frac{π}{2}$

Equation	Solutions
$sin θ = sin α$	$θ = n π if n is odd, if n is even, + (- 1)^{n} α ⇓ θ = (2 m + 1) π - α θ = 2 m π + α$
$cos θ = cos α$	$θ = 2 n π \pm α$
$tan θ = tan α$	$θ = n π + α$

a Gather all the trigonometric functions and constants on different sides of the equation and simplify.

b Use the Double-Angle Identity for cosine to write

cos 2 θ

in terms of

sin θ .

Then gather all the terms on one side of the equation and rewrite them as a product of two expressions.

Solution

a To solve the given equation, start by rewriting it so that all the trigonometric function appear on one side of the equation and the constants on the other side.

4 tan θ = 3 + tan θ

SubEqn

$LHS - tan θ = RHS - tan θ$

3 tan θ = 3

DivEqn

$LHS / 3 = RHS / 3$

tan θ = 1

Now, try to find the angle

θ

for which the value of

tan θ

equals

1 .

The table of the trigonometric values of notable angles can be useful here.

θ = 4 5^{\circ} and θ = 22 5^{\circ}

It can be concluded that this equation has two possible solutions.

b Start by analyzing the given equation.

cos 2 θ = 8 - 15 sin θ

There are two different trigonometric functions involved in the equation, sine and cosine. Therefore, the first step is to rewrite one side of the equation in terms of the other. To do this, the Double-Angle Identity for cosine can be used.

cos 2 θ = 8 - 15 sin θ

$cos (2 θ) = 1 - 2 sin^{2} (θ)$

1 - 2 sin^{2} θ = 8 - 15 sin θ

AddEqn

$LHS + 2 sin^{2} θ = RHS + 2 sin^{2} θ$

1 = 2 sin^{2} θ + 8 - 15 sin θ

SubEqn

$LHS - 1 = RHS - 1$

0 = 2 sin^{2} θ + 7 - 15 sin θ

RearrangeEqn

Rearrange equation

2 sin^{2} θ + 7 - 15 sin θ = 0

CommutativePropAdd

Commutative Property of Addition

2 sin^{2} θ - 15 sin θ + 7 = 0

Next, to solve this equation, rewrite the left-hand side as a product of two expressions.

2 sin^{2} θ - 15 s i n θ + 7 = 0

Rewrite

Rewrite $15 sin θ$ as $14 sin θ + sin θ$

2 sin^{2} θ - (14 s i n θ + s i n θ) + 7 = 0

Distr

Distribute $- 1$

2 sin^{2} θ - 14 sin θ - sin θ + 7 = 0

▼

Factor

FactorOut

Factor out $2 sin θ$

2 sin θ (sin θ - 7) - sin θ + 7 = 0

FactorOut

Factor out $- 1$

2 sin θ (sin θ - 7) - (sin θ - 7) = 0

FactorOut

Factor out $(sin θ - 7)$

(2 sin θ - 1) (sin θ - 7) = 0

By the Zero Product Property, in order for the equation to be true, at least one of the parentheses should be equal to

0 .

(2 sin θ - 1) (sin θ - 7) = 0 ⇓ I . 2 sin θ - 1 = 0 or II . sin θ - 7 = 0

Note that adding

7

to both sides of the Equation

(II)

results in

sin θ = 7 .

However, this is not possible, as the values of sine are always between

- 1

and

1 .

Therefore, this equation has no solution. Now consider Equation

(I) .

2 sin θ - 1 = 0

AddEqn

$LHS + 1 = RHS + 1$

2 sin θ = 1

DivEqn

$LHS / 2 = RHS / 2$

sin θ = \frac{1}{2}

$sin^{- 1} (LHS) = sin^{- 1} (RHS)$

sin^{- 1} (sin θ) = sin^{- 1} (\frac{1}{2})

$f^{- 1} (f (x)) = x$

θ = sin^{- 1} (\frac{1}{2})

To find the value of

sin^{- 1} (\frac{1}{2}),

or arcsine of

\frac{1}{2},

think about the angles for which the value of sine is

\frac{1}{2} .

According to the table of the trigonometric ratios of common angles, there are two such angles in the interval between

0^{\circ}

and

36 0^{\circ} .

Therefore, the given equation has two possible solutions.

θ = 3 0^{\circ} and θ = 15 0^{\circ}

Now that all the solutions have been found, the code mentioned at the beginning can be determined. In Part A, the solutions

45

and

225

were found, while the solutions from Part B were

30

and

150 .

The code can be found by combining these numbers in the order directed at the start.

4522530150

Try it out to see the hidden message!

a Use the Double-Angle Identity for cosine to rewrite

cos 2 θ

in terms of

cos θ .

b Apply one of the Pythagorean Identities and simplify the equation.

Solution

a In the first equation, the same trigonometric function appears two times but it has different arguments. Therefore, use the Double-Angle Identity for cosine to rewrite the equation.

cos 2 θ = 2 cos^{2} θ - 1

Apply this identity and then simplify the equation.

cos 2 θ + 4 cos θ = - 3

$cos (2 θ) = 2 cos^{2} (θ) - 1$

2 cos^{2} θ - 1 + 4 cos θ = - 3

AddEqn

$LHS + 3 = RHS + 3$

2 cos^{2} θ + 2 + 4 cos θ = 0

CommutativePropAdd

Commutative Property of Addition

2 cos^{2} θ + 4 cos θ + 2 = 0

DivEqn

$LHS / 2 = RHS / 2$

cos^{2} θ + 2 cos θ + 1 = 0

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

(cos θ + 1)^{2} = 0

SqrtEqn

$LHS = RHS$

cos θ + 1 = 0

SubEqn

$LHS - 1 = RHS - 1$

cos θ = - 1

In order to find the values of

θ

that satisfy this equation, think about the angles for which the cosine equals

- 1 .

According to the table of the trigonometric ratios of common angles, there is at least one such angle.

θ = π

Since cosine is a periodic function with a period of

2 π

, for every integer

n,

the following angles are the solutions to the equation.

θ = π + 2 π n

b Start by analyzing the given equation.

3 tan^{2} θ - 1 = sec^{2} θ

There are two trigonometric functions in the equation, tangent and secant. Recall that one of the Pythagorean Identities relates these two functions.

1 + tan^{2} θ = sec^{2} θ

Substitute

1 + tan^{2} θ

for

sec^{2} θ

into the equation and simplify it.

3 tan^{2} θ - 1 = sec^{2} θ

Substitute

$sec^{2} θ = 1 + t a n^{2} θ$

3 tan^{2} θ - 1 = 1 + t a n^{2} θ

SubEqn

$LHS - tan^{2} θ = RHS - tan^{2} θ$

2 tan^{2} θ - 1 = 1

AddEqn

$LHS + 1 = RHS + 1$

2 tan^{2} θ = 2

DivEqn

$LHS / 2 = RHS / 2$

tan^{2} θ = 1

SqrtEqn

$LHS = RHS$

tan θ = \pm 1

Next, look for the angles for which tangent is equal to

1

- 1 .

θ θ = 4 5^{\circ} or \frac{π}{4} = 22 5^{\circ} or \frac{5 π}{4} θ = 13 5^{\circ} or \frac{3 π}{4} θ = 31 5^{\circ} or \frac{7 π}{4}

To find the general equation for the solutions, note that all these angles can be expressed in the following way.

\frac{π}{4} \frac{3 π}{4} \frac{5 π}{4} \frac{7 π}{4} = \frac{π}{4} + \frac{π}{2} \cdot 0 = \frac{π}{4} + \frac{π}{2} \cdot 1 = \frac{π}{4} + \frac{π}{2} \cdot 2 = \frac{π}{4} + \frac{π}{2} \cdot 3

Therefore, all the solutions to the equation can be given by the following equation where

n

is an integer number.

θ = \frac{π}{4} + \frac{π}{2} n

After correctly solving both equations, Magdalena can finally read the joke that Davontay found for her.

a Use the Pythagorean Identity to rewrite

cos θ

in terms of

sin θ .

Then factor the equation and apply the Zero Product Property.

b Move

cos θ

to the right side of the equation and then square both sides. Check the solutions by substituting them into the original equation and seeing if it remains true.

a Start by examining the first equation.

2 cos^{2} θ + 3 sin θ = 3

Since there are two different trigonometric functions, one should be rewritten in terms of the other. To do so, use the Pythagorean Identity, but first rewrite the equation so that

cos^{2} θ

is isolated on one side of the equation.

sin^{2} θ + cos^{2} θ = 1 ⇓ cos^{2} θ = 1 - sin^{2} θ

Substitute

1 - sin^{2} θ

for

cos^{2} θ

and simplify the equation.

2 cos^{2} θ + 3 sin θ = 3

Substitute

$cos^{2} θ = 1 - s i n^{2} θ$

2 (1 - s i n^{2} θ) + 3 sin θ = 3

Distr

Distribute $2$

2 - 2 sin^{2} θ + 3 sin θ = 3

CommutativePropAdd

Commutative Property of Addition

- 2 sin^{2} θ + 3 sin θ + 2 = 3

SubEqn

$LHS - 3 = RHS - 3$

- 2 sin^{2} θ + 3 sin θ - 1 = 0

MultEqn

$LHS \cdot (- 1) = RHS \cdot (- 1)$

2 sin^{2} θ - 3 sin θ + 1 = 0

Next, factor the expression on the left-hand side into two parentheses.

2 sin^{2} θ - 3 sin θ + 1 = 0

Rewrite

Rewrite $3 sin θ$ as $2 sin θ + sin θ$

2 sin^{2} θ - (2 sin θ + sin θ) + 1 = 0

Distr

Distribute $- 1$

2 sin^{2} θ - 2 sin θ - sin θ + 1 = 0

FactorOut

Factor out $2 sin θ$

2 sin θ (sin θ - 1) - sin θ + 1 = 0

FactorOut

Factor out $- 1$

2 sin θ (sin θ - 1) - (sin θ - 1) = 0

FactorOut

Factor out $sin θ - 1$

(2 sin θ - 1) (sin θ - 1) = 0

By the Zero Product Property, at least one of the parentheses must be equal to

0 .

Two equations can be formed by applying this property.

2 sin θ - 1 = 0 sin θ - 1 = 0 (I) (II)

$(I), (II):$ $LHS + 1 = RHS + 1$

2 sin θ = 1 sin θ = 1

DivEqn

$(I):$ $LHS / 2 = RHS / 2$

sin θ = \frac{1}{2} sin θ = 1

To solve these two equations, try to find the angles for which the values of sine are

\frac{1}{2}

and

1 .

For the first equation, there are two such angles —

\frac{π}{6}

and

\frac{5 π}{6}

— while for the second there is only one —

\frac{π}{2} .

θ_{1} = \frac{π}{6}, θ_{2} = \frac{5 π}{6}, θ_{3} = \frac{π}{2}

Finally, the solutions should be checked whether there are any extraneous solutions. To do that, substitute each of them into the original equation and see if it is true.

2 cos^{2} θ + 3 sin θ = 3

Substitute

$θ = \frac{π}{6}$

2 cos^{2} (\frac{π}{6}) + 3 sin (\frac{π}{6}) = ? 3

▼

Simplify left-hand side

2 (\frac{3}{2})^{2} + 3 sin (\frac{π}{6}) = ? 3

2 (\frac{3}{2})^{2} + 3 (\frac{1}{2}) = ? 3

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

2 \frac{3}{4} + 3 (\frac{1}{2}) = ? 3

MoveLeftFacToNum

$a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$

\frac{6}{4} + \frac{3}{2} = ? 3

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

\frac{3}{2} + \frac{3}{2} = ? 3

AddFrac

Add fractions

3 = 3 ✓

The other two solutions can be checked in a similar fashion.

Solution	Substitute	Evaluate	True or False
$θ = \frac{π}{6}$	$2 cos^{2} (\frac{π}{6}) + 3 sin (\frac{π}{6}) = ? 3$	$2 (\frac{3}{2})^{2} + 3 (\frac{1}{2}) = ? 3$	$3 = 3 ✓$
$θ = \frac{5 π}{6}$	$2 cos^{2} (\frac{5 π}{6}) + 3 sin (\frac{5 π}{6}) = ? 3$	$2 (- \frac{3}{2})^{2} + 3 (\frac{1}{2}) = ? 3$	$3 = 3 ✓$
$θ = \frac{π}{2}$	$2 cos^{2} (\frac{π}{2}) + 3 sin (\frac{π}{2}) = ? 3$	$2 (0)^{2} + 3 (1) = ? 3$	$3 = 3 ✓$

Therefore, there are three solutions to the equation and none of them are extraneous.

b Again, begin by analyzing the given equation.

sin θ - cos θ = 0

First, move

cos θ

to the right-hand side of the equation. Then, square each side of the equation.

sin θ = cos θ ⇓ sin^{2} θ = cos^{2} θ

Now, to rewrite one trigonometric function in terms of the other, use the Pythagorean Identity. Similarly to Part A, substitute

1 - sin^{2} θ

for

cos^{2} θ .

sin^{2} θ = cos^{2} θ

Substitute

$cos^{2} θ = 1 - s i n^{2} θ$

sin^{2} θ = 1 - s i n^{2} θ

AddEqn

$LHS + sin^{2} θ = RHS + sin^{2} θ$

2 sin^{2} θ = 1

DivEqn

$LHS / 2 = RHS / 2$

sin^{2} θ = \frac{1}{2}

SqrtEqn

$LHS = RHS$

sin θ = \pm \frac{1}{2}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

sin θ = \pm \frac{1}{2}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

sin θ = \pm \frac{2}{2}

To find the solutions of this equation, look for the angles for which the sine equals

\pm \frac{2}{2} .

The table with the trigonometric ratios of notable angles can be useful here.

θ_{1} = \frac{π}{4} θ_{2} = \frac{3 π}{4} θ_{3} = \frac{5 π}{4} θ_{4} = \frac{7 π}{4}

Finally, substitute each of these solutions into the original equation in order to determine whether any of them is extraneous.

Solution	Substitute	Evaluate	True or False
$θ_{1} = \frac{π}{4}$	$sin (\frac{π}{4}) - cos (\frac{π}{4}) = ? 0$	$\frac{2}{2} - \frac{2}{2} = ? 0$	$0 = 0 ✓$
$θ_{2} = \frac{3 π}{4}$	$sin (\frac{3 π}{4}) - cos (\frac{3 π}{4}) = ? 0$	$\frac{2}{2} - (- \frac{2}{2}) = ? 0$	$2 \neq = 0 \times$
$θ_{3} = \frac{5 π}{4}$	$sin (\frac{5 π}{4}) - cos (\frac{5 π}{4}) = ? 0$	$- \frac{2}{2} - (- \frac{2}{2}) =$

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Catch-Up and Review

Riding a Ferris Wheel

Presenting Inverse Trigonometric Functions

Inverse Trigonometric Functions

Trigonometric Equations and Their Solutions

Trigonometric Equation

Decoding the Message with the Solutions to Trigonometric Equations

Hint

Solution

Math Challenge With a Gift

Hint

Solution

Making a Bet on Equations

Hint

Solution

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