top of page

 

7.6 Solving Quadratic Trigonometric Equations

Helpful Videos

Recall that not all trigonometric equations are identities and, similar to when we solve linear trigonometric equations, the goal is to solve for the unknown values of cos, sin or any of the variables with unknown values.

Sample Equations

sin^2(x) + sin(x) -1 = 0

tan^2(x)sin(x) = 2tan(x)

5cos^2(x) + 3cos(x) + 4

tan^2(x) -sin(x)

Example 1

For 0 < x < 2π Solve for 2sin^2(x) + sin(x) = 1

Step 1) Rearrange so that the trigonometric equation equals 0

 

Step 2) If you want think of the equation as a quadratic and replace the sin, cos, tan ect as X

 

Step 3) Solve like a quadratic equation and find the factors

 

Step 4) Now sub your variable back into the equation set each braket to zero and solve for the variable.

 

Step 5) Now that you have values using your perameters in this case 0 < x < 2π using the unit circle find where sin = x or cos = x

 

Step 6) In this case find where sin(x) = 1/2 and where sin(x)= -1 where 0 < x < 2π

 2sin^2(x) + sin(x) - 1 = 0

2x^2 + x -1 = 0

(2x-1) (x+1) = 0

 2sin^2(x) + sin(x) - 1 = 0

(2x-1) = 0

2sin(x)-1= 0

2sin(x) = 1

sin(x) = 1/2

(x + 1) = 0

sin(x)+1= 0

sin(x) = -1

 

sin(x) = 1/2 sin(x) = 30 ,150 sin(x) = π/6 or 5π/6

sin(x) = -1 sin(x) = 270 sin(x) = 3π/2

For 0 < x < 2π Solve for 2cos^2(x) + 3sin(x) -3 = 0

Example 2

Step 1) If there are multiple trigonometric functions in an equation identities may need to be used before factoring. 

 

Step 2) Replace cos^2x as (1-sin^2x ) because of the identity sin^2x + cos^x = 1

 

Step 3) Rearrange into a quadratic equation and solve

 

Step 4) Now sub your variable back into the equation set each braket to zero and solve for the variable.

 

Step 5) Now that you have values using your perameters in this case 0 < x < 2π using the unit circle find where sin = x or cos = x

 

Step 6) In this case find where sin(x) = 1/2 and where sin(x)= -1 where 0 < x < 2π

 2cos^2(x) + 3sin(x) - 3 = 0

-2sin^2(x) + 3sin(x) -3 +2 = 0

-2sin^2(x) + 3sin(x) -1 = 0

(2x - 1)(x - 1) = 0

(2sin(x) - 1)(sin(x) - 1) = 0

 2(1-sin^2x ) + 3sin(x) - 3 = 0

(2x-1) = 0

2sin(x)-1= 0

2sin(x) = 1

sin(x) = 1/2

(x - 1) = 0

sin(x)-1= 0

sin(x) = 1

 

sin(x) = 1/2 sin(x) = 30 ,150 sin(x) = π/6 or 5π/6

sin(x) = 1 sin(x) = 90 sin(x) = π/2

-2x^2 +3x - 1 = 0

For 0 < x < 2π Solve for (√2sin(x)-1)(2cos(x)+√) = 0

Example 3

Step 1) For this equation you do not need to factor because it is already in factored form. Like the previous examples, set each bracket to zero and solve for sin(x).

 

Step 5) Now that you have values, using your perameters in this case 0 < x < 2π using the unit circle find where sin = x or cos = x

 

Step 6) Be mindful of the signs of the values because this will dictate what quadrants they will be in. For example cos(x) = -√3/2 this means that the value for cos will be negative or in the tan/sin quadrant or quadrants 2 & 3.

(√2sin(x)-1)(2cos(x)+√) = 0

(√2sin(x)-1)= 0

√2sin(x)-1= 0

√2sin(x) = 1

sin(x) = 1/√2

(2cos(x)+√3)= 0

2cos(x)+√3= 0 2cos(x) = -√3

cos(x) = -√3/2

sin(x) = 1/√2 sin(x) = 45, 135 sin(x) = π/4 or 3π/4

cos(x) = -√3/2 cos(x) = 150, 210 cos(x) = 5π/6 or 7π/6

Sample Questions
Level 1/2

A) For 0 < x < 2π Solve for sin^2(x) - sin(x) = 0

B) For 0 < x < 2π Solve for 81cos^2(x) - 64 = 0

C) For 0 < x < 2π Solve for sin(x)cos(x) = 0

Level 3

A) For 0 < x < 2π  tan^2(x) + 4tan(x) - 12 = 0

Level 4(Application)

There is a oceanographer who is studying a phenomena that is studying an area of waves in the ocean for 2π kilometers looking on a hill above sea level he models this equation as 9sin^2(d) - 1 where d is the height of waves. At what interval are the waves at a height of 0.

Key Concepts/Tips

Think of quadratic form ax^2 + bx + c =0

Replace Cos, Sin,Tan...ect as X

Use Quadratic formula for Trigonometric equations that are tricky to factor. 

You can get impossible results Ex. Sin= 2.33 because Sin can only be values between 1 and -1. 

The trigonometric quadratic equations may have more than one answer although some may not be plausible due to the nature of the problem or the nature of trigonometric functions

In some cases there may be multiple trig functions in one equation so to simplify into a single quadratic equation use trigonometric identities or equivalent equations such as double angle formulae, compound angle formula or Pyhtagorean identities to solve the equation.

Some equations may not be able to be factored so using the quadratic equation you can get values for the variables using this method.

bottom of page