7.06 Trigonometric equations with identities

We can apply our knowledge of trigonometric identities to manipulate expressions in equations and arrive at more elegant solution paths. Let's review the different trig identities, and how we might apply them in the context of solving trig equations.

Solving trigonometric equations using polynomial identities

Recall that we can rewrite trig expressions by factoring them, or using substitution and different polynomial identities. We can apply these techniques and the zero product property to solve trig equations that mimic polynomial equations.

 The following polynomial identities also apply to trigonometric expressions of the same form:

	Polynomial Identity
Distributive property / Greatest common factor (GCF)	$AB+AC+\dots=A\left(B+C+\dots\right)$AB+AC+…=A(B+C+…)
Perfect square trinomial	$A^2+2AB+B^2=\left(A+B\right)^2$A2+2AB+B2=(A+B)2 $A^2-2AB+B^2=\left(A-B\right)^2$A2−2AB+B2=(A−B)2
Difference of squares	$A^2-B^2=\left(A+B\right)\left(A-B\right)$A2−B2=(A+B)(A−B)
Sum of cubes	$A^3+B^3=\left(A+B\right)\left(A^2-AB+B^2\right)$A3+B3=(A+B)(A2−AB+B2)
Difference of cubes	$A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)$A3−B3=(A−B)(A2+AB+B2)
Factoring quadratics (leading coefficient of 1)	$x^2+\left(A+B\right)x+AB=\left(x+A\right)\left(x+B\right)$x2+(A+B)x+AB=(x+A)(x+B)
Distributive property / Grouping in pairs	$\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD
Quadratic formula	If $Ax^2+Bx+C=0$Ax2+Bx+C=0 then $x=\frac{-B\pm\sqrt{B^2-4AC}}{2A}$x=−B±√B2−4AC2A​
Zero product property	The equation $AB=0$AB=0 is true if and only if $A=0$A=0 or $B=0$B=0.

The key when facing questions involving these techniques is to figure out which to use and when.

 

Worked examples

Question 1

Solve $\cot x\cos^2x=2\cot x$cotxcos2x=2cotx 

Think: In order to see the trigonometric equation as a quadratic equation, we can substitute the $\cot x=A$cotx=A and $\cos x=B$cosx=B . This gives us the following equation.

$AB^2=2A$AB2=2A

Do: 

$\cot x\cos^2x$cotxcos2x	$=$=	$2\cot x$2cotx	Given
$AB^2$AB2	$=$=	$2A$2A	Let $\cot x=A$cotx=A and $\cos x=B$cosx=B
$AB^2-2A$AB2−2A	$=$=	$0$0	Subtract $2A$2A from both sides
$A\left(B^2-2\right)$A(B2−2)	$=$=	$0$0	Factor out the greatest common factor of $A$A

 

Now, let's substitute the trigonometric functions back into the equation.

$A\left(B^2-2\right)$A(B2−2)	$=$=	$0$0
$\cot x(\cos^2x-2)$cotx(cos2x−2)	$=$=	$0$0

Set each factor equal to zero using the zero product property.

Using the unit circle, we know the angle $\frac{\pi}{2}$π2​ radians will produce $\tan\frac{\pi}{2}=\frac{1}{0}$tanπ2​=10​ . There is no angle on the unit circle that will produce $\cos x=\pm\sqrt{2}$cosx=±√2. Using the graphing calculator, we discover that there is no solution for $\cos x=\pm\sqrt{2}$cosx=±√2 because $\pm\sqrt{2}$±√2 is outside the range of the $\cos$cos function.   

Answer: $x=\frac{\pi}{2}$x=π2​ radians

Reflect: How would we write the solution so it represents all possible values of $x$x? Using the graphing calculator to find the period, we determine that the period of $\cot x\cos^2x=2\cot x$cotxcos2x=2cotx is $\pi$π. Remember, a period of a function is the interval of $x$x-values where the graph repeated in both directions along the $x$x-axis. The general form of the solution is found by adding multiples of $\pi$π to $x=\frac{\pi}{2}$x=π2​.

$x=\frac{\pi}{2}+k\pi$x=π2​+kπ

where $k$k is an integer.    

Question 2

Find $x$x in radians such that $\tan^2x-2\tan x=3$tan2x−2tanx=3. Round to the 3 decimal places.

Think: In order to see the trigonometric equation as a quadratic equation, we can substitute the $\tan x=A$tanx=A.

$A^2-2*A=3$A2−2*A=3

Do: 

$\tan^2x-2\tan x=3$tan2x−2tanx=3	$=$=	$3$3	Given
$A^2-2A$A2−2A	$=$=	$3$3	Substitute $\tan x=A$tanx=A
$A^2-2A-3$A2−2A−3	$=$=	$0$0	Subtract $3$3 from both sides
$\left(A-3\right)\left(A+1\right)$(A−3)(A+1)	$=$=	$0$0	Factor the quadratic expression

 

Now, let's substitute the trigonometric functions back into the equation.

$\left(A-3\right)\left(A+1\right)$(A−3)(A+1)	$=$=	$0$0
$(\tan x-3)(\tan x+1)$(tanx−3)(tanx+1)	$=$=	$0$0

Set each factor equal to zero using the zero product property.

Using the unit circle, we know the angle $-\frac{\pi}{4}$−π4​ radians will produce$\tan-\frac{\pi}{4}=-1$tan−π4​=−1 . There is no angle on the unit circle that will produce $\tan x=3$tanx=3. Using the graphic calculator, we discover that $\tan^{-1}3=1.249$tan−13=1.249 radians.   

Answer: $x=-\frac{\pi}{4}$x=−π4​ and $x=1.249$x=1.249 radians

Reflect: How would we write the solution so it represents all possible values of $x$x? Using the graphic calculator to find the period, we determine that the period of $\tan^2x-2\tan x-3=0$tan2x−2tanx−3=0 is $\pi$π.  The general form of the solution is found by adding multiples of $\pi$π. 

$x=-\frac{\pi}{4}+k\pi$x=−π4​+kπ and $x=1.249+k\pi$x=1.249+kπ

where $k$k is an integer.    

 

Solving trigonometric equations using trig identities

We can also apply our knowledge of trigonometric identities to rewrite trig expressions and solve trig equations. Let's review the identities and work through some examples of how we might use them to solve a trig equation.

Worked examples

Question 3

Find the value of $x$x in radians for the equation $\sin^2x+\cos^2x+\cos x=\frac{\sqrt{3}}{2}+1$sin2x+cos2x+cosx=√32​+1.

Think: Does the equation contain different trigonometric functions? Yes. Can one or more functions be modified using the reciprocal functions ($\csc$csc, $\sec$sec, and $\cot$cot) so that the equation contains only one trigonometric function? No, the $\sin$sin function cannot be changed into $\cos$cos. So, we must review the trigonometric identities listed above. Is there an identity that we could use? Yes.

$\sin^2x+\cos^2x=1$sin2x+cos2x=1 

 

Do: Use the identity to substitute the value of $\sin^2x+\cos^2x=1$sin2x+cos2x=1 :

$\sin^2x+\cos^2x+\cos x$sin2x+cos2x+cosx	$=$=	$\frac{\sqrt{3}}{2}+1$√32​+1	Given
$(\sin^2x+\cos^2x)+\cos x$(sin2x+cos2x)+cosx	$=$=	$\frac{\sqrt{3}}{2}+1$√32​+1	Group the terms that form one side of the identity: $\sin^2x+\cos^2x=1$sin2x+cos2x=1
$(1)+\cos x$(1)+cosx	$=$=	$\frac{\sqrt{3}}{2}+1$√32​+1	Substitute the grouped terms for $1$1
$\cos x$cosx	$=$=	$\frac{\sqrt{3}}{2}$√32​	Subtract $1$1 from both sides
$x$x	$=$=	$\cos^{-1}\frac{\sqrt{3}}{2}$cos−1√32​

Using the unit circle, we know that the angle $\frac{\pi}{6}$π6​ radians produces the $\cos x=\frac{\sqrt{3}}{2}$cosx=√32​ within the first quadrant. 

Reflect: If trigonometric functions repeat over a period, then we can write the solution in general form to represent all possible solutions. First, we must find the period. Using the graphic calculator, we notice that the period of the function is $2\pi$2π.  Finally, by adding multiples of $2\pi$2π we obtain the general solution

$x=\frac{\pi}{6}+2\pi k$x=π6​+2πk

Question 4

Find the value of $\alpha$α in degrees for $\sec\alpha=\csc(\frac{\alpha}{2}+20^\circ)$secα=csc(α2​+20°). Write answer in fractional form.

Think: Does the equation contain different trigonometric functions? Yes. Can one or more functions be modified using the reciprocal functions ($\csc$csc, $\sec$sec, and $\cot$cot) so that the equation contains only one trigonometric function? No, the $\sec$sec function cannot be changed into $\csc$csc. So, we must review the trigonometric identities listed above. Is there an identity that we could use? Yes.

$\sec\alpha=\csc(90^\circ-\alpha)$secα=csc(90°−α)

Do: Use the identity to substitute the value of $\sec\alpha=\csc(90^\circ-\alpha)$secα=csc(90°−α) :

$\sec\alpha$secα	$=$=	$\csc(\frac{\alpha}{2}+20^\circ)$csc(α2​+20°)	Given
$\csc(90^\circ-\alpha)$csc(90°−α)	$=$=	$\csc(\frac{\alpha}{2}+20^\circ)$csc(α2​+20°)	Substitute $\sec\alpha$secα for $\csc(90^\circ-\alpha)$csc(90°−α)
$(90^\circ-\alpha)$(90°−α)	$=$=	$(\frac{\alpha}{2}+20^\circ)$(α2​+20°)	Since both sides are $\csc$csc functions, set the angles equal
$-\frac{3\alpha}{2}$−3α2​	$=$=	$-70$−70	Subtract $90^\circ$90° and $\frac{\alpha}{2}$α2​ from both sides
$\alpha$α	$=$=	$\frac{140}{3}$1403​	Multiply both sides by $-\frac{2}{3}$−23​

 

question 5

Find the solutions in the interval $[0,2\pi)$[0,2π) for $\sin(x+\frac{\pi}{4})+\sin(x-\frac{\pi}{4})=-1$sin(x+π4​)+sin(x−π4​)=−1 in radians.

Think: Does the equation contain different trigonometric functions? Yes. There are two $\sin$sin  functions; however, the angles are different. Similar to combining like terms with unlike variables, we cannot combine trigonometric functions with unlike angles. Notice, the angles are expressions. So, we must review the trigonometric identities listed above. Is there an identity or identities that we could use? Yes.

$\sin\left(A+B\right)=\sin A\cos B+\cos A\sin B$sin(A+B)=sinAcosB+cosAsinB

$\sin\left(A-B\right)=\sin A\cos B-\cos A\sin B$sin(A−B)=sinAcosB−cosAsinB

Do: Use the identity to substitute:

$\sin(x+\frac{\pi}{4})+\sin(x-\frac{\pi}{4})$sin(x+π4​)+sin(x−π4​)	$=$=	$-1$−1	Given
$(\sin x\cos\frac{\pi}{4}+\cos x\sin\frac{\pi}{4})+(\sin x\cos\frac{\pi}{4}-\cos x\sin\frac{\pi}{4})$(sinxcosπ4​+cosxsinπ4​)+(sinxcosπ4​−cosxsinπ4​)	$=$=	$-1$−1	Substitute both identities into the equation
$2\sin x\cos\frac{\pi}{4}$2sinxcosπ4​	$=$=	$-1$−1	Combine like terms
$2\sin x\frac{\sqrt{2}}{2}$2sinx√22​	$=$=	$-1$−1	Substitute the value of $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$cosπ4​=√22​
$\sqrt{2}\sin x$√2sinx	$=$=	$-1$−1	Combine like terms
$\sin x$sinx	$=$=	$-\frac{1}{\sqrt{2}}$−1√2​	Divide both sides by $\sqrt{2}$√2
$\sin x$sinx	$=$=	$-\frac{\sqrt{2}}{2}$−√22​	Rationalize the denominator

Using the unit circle, we find that there are only $2$2 solutions in the interval $[0,2\pi)$[0,2π):

$x=\frac{5\pi}{4}$x=5π4​

$x=\frac{7\pi}{4}$x=7π4​

 

By letting $B=2A$B=2A we obtain three corresponding half-angle identities. These identities are sometimes referred to in alternate forms which have been manipulated by applying the Pythagorean identities. The relationships remain the same.

question 6

Find all solutions of $2\cos x+\sin2x=0$2cosx+sin2x=0 in the interval $[0,2\pi)$[0,2π).

Think: Does the equation contain different trigonometric functions? Yes. Can one or more functions be modified using the reciprocal functions ($\csc$csc, $\sec$sec, and $\cot$cot) so that the equation contains only one trigonometric function? No, the $\sin$sin function cannot be changed into $\cos$cos. So, we must review the trigonometric identities listed above. Is there an identity that we could use? Yes. Because the $\sin$sin function has an angle that is a being doubled, then we can use the double-angle formula.

$\sin2A=2\sin A\cos A$sin2A=2sinAcosA

Do: Use the identity to substitute the value of $\sin2A=2\sin A\cos A$sin2A=2sinAcosA:

$2\cos x+\sin2x$2cosx+sin2x	$=$=	$0$0	Given
$2\cos x+2\sin x\cos x$2cosx+2sinxcosx	$=$=	$0$0	Substitute $\sin2x=2\sin x\cos x$sin2x=2sinxcosx
$2\cos x(1+\sin x$2cosx(1+sinx	$=$=	$0$0	Factor out $2\cos x$2cosx
$2\cos x=0$2cosx=0		$1+\sin x=0$1+sinx=0	Set factors equal to zero
$\cos x=0$cosx=0		$\sin x=-1$sinx=−1	Isolate the trigonometric function

Using the unit circle, we find that $\cos x=0$cosx=0  has $2$2 solutions in the interval$[0,2\pi)$[0,2π):

$x=\frac{\pi}{2}$x=π2​

$x=\frac{3\pi}{2}$x=3π2​

and $\sin x=-1$sinx=−1 has $1$1 solution:

$x=\frac{3\pi}{2}$x=3π2​

Therefore, the final solution within in interval $[0,2\pi)$[0,2π) is $x=\frac{\pi}{2}$x=π2​ and $x=\frac{3\pi}{2}$x=3π2​

Practice questions

questions 7

Question 8

Question 9