Quadratic equations in sin(θ), cos(θ) or tan(θ) can be solved by first solving the quadratic in an arbitrary variable, then solving the simple trigonometric equations corresponding to the roots. More generally, you can use the trigonometric identities sin2(θ)+cos2(θ)≡1 and cos(θ)sin(θ)≡tan(θ) to help you solve a wide range of equations involving multiple trigonometric functions.
Quadratic equations in a single trigonometric function
Suppose you are given an equation in one of the following forms:
Where a, b and c are constants. The following procedure allows you to find all solutions to any of the above equations in any interval D≤θ≤D′:
PROCEDURE
1.
Factorise the quadratic equation ax2+bx+c into its linear roots to get:
ax2+bx+c=a(x−α+)(x−α−)
where α+=2a−b+b2−4ac and α−=2a−b−b2−4ac
2.
Find all solutions to either sin(θ)=α+, cos(θ)=α+ or tan(θ)=α+ in the interval D≤θ≤D′, depending on which trigonometric function appears in your equation.
3.
Find all solutions to either sin(θ)=α−, cos(θ)=α− or tan(θ)=α− in the interval D≤θ≤D′, depending on which trigonometric function appears in your equation.
Example 1
Find all solutions to the equation sin2(θ)=41in the interval 0°≤θ≤360°.
Consider the equation x2=41. Rearrange and factorise to get:
x2x2−41(x−21)(x+21)=41=0=0
This means that the equation sin2(θ)=41 can be solved by solving the two equations:
sin(x)=21
sin(x)=−21
The principal value for the first root is sin−1(21)°=30°. Using acute angle identities or otherwise, determine that the only solutions to sin(θ)=21 in the interval 0°≤θ≤360° are θ=30°,150°.
The principal value for the second root is sin−1(−21)°=−30°. Using acute angle identities or otherwise, determine that the only solutions to sin(θ)=−21 in the interval 0°≤θ≤360° are θ=210°,θ=330°.
Therefore, the solutions to the equation sin2(θ)=41 in the interval 0°≤θ≤360° are θ=30°, θ=150°, θ=210° and θ=330°.
Equations in multiple trigonometric functions
You will often be given equations with multiple different trigonometric functions. Using the identities sin2(θ)+cos2(θ)≡1 and cos(θ)sin(θ)≡tan(θ), you will need to rearrange and factorise these equations into simple trigonometric equations you can solve.
Most of the time it will be possible to use these identities to transform the equation into a quadratic equation in a single trigonometric function, which can be solved by applying the above procedure.
Example 2
Find all solutions to the equation cos(θ)(cos(θ)+tan(θ))=0 in the interval 0°≤θ≤360°. Give your answers to 3s.f.
Use the identities tan(θ)≡cos(θ)sin(θ) and sin2(θ)+cos2(θ)≡1 to rearrange the equation so that it is a quadratic in sin(θ):
Therefore, it suffices to find all solutions to the equation sin2(θ)−sin(θ)−1=0 in the interval 0°≤θ≤360°.
The equation x2−x−1=0 yields the solutions x=21±5.
The equation sin(θ)=21+5 has no solutions, as 21+5>1.
The principal value for the second root is sin−1([21−5])°=−38.17…°. Using acute angle identities or otherwise, determine that the only solutions to the equation sin(θ)=21−5 in the interval 0°≤θ≤360° are θ=218.1…°,321.8…°.
Therefore, to 3s.f., the solutions to cos(θ)(cos(θ)+tan(θ))=0 in the interval 0°≤θ≤360° are θ=218° and θ=322°.
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FAQs - Frequently Asked Questions
How do I solve quadratics in trigonometric functions?
Quadratic equations in trigonometric functions can be solved by first solving the quadratic in an arbitrary variable, then solving the simple trigonometric equations corresponding to the roots.
How do I solve trigonometric equations with identities?
Use identities to get the equation in terms of a single trigonometric function you can solve for.
