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Solving a trigonometric equation in quadratic form by factoring

Solve the equation exactly: 2 sin 2 θ 5 sin θ + 3 = 0 , 0 θ 2 π .

Using grouping, this quadratic can be factored. Either make the real substitution, sin θ = u , or imagine it, as we factor:

  2 sin 2 θ 5 sin θ + 3 = 0 ( 2 sin θ 3 ) ( sin θ 1 ) = 0

Now set each factor equal to zero.

2 sin θ 3 = 0 2 sin θ = 3 sin θ = 3 2 sin θ 1 = 0 sin θ = 1

Next solve for θ : sin θ 3 2 , as the range of the sine function is [ −1 , 1 ] . However, sin θ = 1 , giving the solution θ = π 2 .

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Solve sin 2 θ = 2 cos θ + 2 , 0 θ 2 π . [Hint: Make a substitution to express the equation only in terms of cosine.]

cos θ = 1 , θ = π

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Solving a trigonometric equation using algebra

Solve exactly:

2 sin 2 θ + sin θ = 0 ; 0 θ < 2 π

This problem should appear familiar as it is similar to a quadratic. Let sin θ = x . The equation becomes 2 x 2 + x = 0. We begin by factoring:

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

Set each factor equal to zero.

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

Then, substitute back into the equation the original expression sin θ for x . Thus,

sin θ = 0 θ = 0 , π sin θ = 1 2 θ = 7 π 6 , 11 π 6

The solutions within the domain 0 θ < 2 π are θ = 0 , π , 7 π 6 , 11 π 6 .

If we prefer not to substitute, we can solve the equation by following the same pattern of factoring and setting each factor equal to zero.

2 sin 2 θ + sin θ = 0 sin θ ( 2 sin θ + 1 ) = 0 sin θ = 0 θ = 0 , π   2 sin θ + 1 = 0 2 sin θ = 1 sin θ = 1 2 θ = 7 π 6 , 11 π 6
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Solving a trigonometric equation quadratic in form

Solve the equation quadratic in form exactly: 2 sin 2 θ 3 sin θ + 1 = 0 , 0 θ < 2 π .

We can factor using grouping. Solution values of θ can be found on the unit circle.

( 2 sin θ 1 ) ( sin θ 1 ) = 0   2 sin θ 1 = 0 sin θ = 1 2 θ = π 6 , 5 π 6 sin θ = 1 θ = π 2
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Solve the quadratic equation 2 cos 2 θ + cos θ = 0.

π 2 , 2 π 3 , 4 π 3 , 3 π 2

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Solving trigonometric equations using fundamental identities

While algebra can be used to solve a number of trigonometric equations, we can also use the fundamental identities because they make solving equations simpler. Remember that the techniques we use for solving are not the same as those for verifying identities. The basic rules of algebra apply here, as opposed to rewriting one side of the identity to match the other side. In the next example, we use two identities to simplify the equation.

Use identities to solve an equation

Use identities to solve exactly the trigonometric equation over the interval 0 x < 2 π .

cos x cos ( 2 x ) + sin x sin ( 2 x ) = 3 2

Notice that the left side of the equation is the difference formula for cosine.

cos x cos ( 2 x ) + sin x sin ( 2 x ) = 3 2 cos ( x 2 x ) = 3 2 Difference formula for cosine cos ( x ) = 3 2 Use the negative angle identity . cos x = 3 2

From the unit circle in [link] , we see that cos x = 3 2 when x = π 6 , 11 π 6 .

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Solving the equation using a double-angle formula

Solve the equation exactly using a double-angle formula: cos ( 2 θ ) = cos θ .

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

cos ( 2 θ ) = cos θ 2 cos 2 θ 1 = cos θ 2 cos 2 θ cos θ 1 = 0 ( 2 cos θ + 1 ) ( cos θ 1 ) = 0 2 cos θ + 1 = 0 cos θ = 1 2 cos θ 1 = 0 cos θ = 1

So, if cos θ = 1 2 , then θ = 2 π 3 ± 2 π k and θ = 4 π 3 ± 2 π k ; if cos θ = 1 , then θ = 0 ± 2 π k .

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Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
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