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Solve trigonometric equations over specified intervals

Solving trigonometric equations means finding every angle in a specified interval, in degrees or radians, that satisfies equations involving sine, cosine, or tangent, including forms simplified through algebraic manipulation, basic identities, factoring, or substitution. The reasoning depends on periodicity, quadrant relationships, inverse-trigonometric values, and careful restriction of general solutions to the interval; solutions must be checked, especially when squaring or dividing can introduce or remove candidates. The scope excludes general nonlinear systems, complex-number solutions, and advanced numerical methods for equations without standard algebraic or trigonometric reductions.

Detailed Explanation: Solve trigonometric equations over specified intervals

To solve a trigonometric equation on a specified interval:

  1. Rewrite or factor the equation until you can identify basic trigonometric values.
  2. Find the reference angle using an inverse trigonometric function.
  3. Use the signs in each quadrant to find all possible angles.
  4. Keep only angles in the given interval.
  5. Check the solutions in the original equation.

Example

Solve

2sin⁡2x−3sin⁡x+1=02\sin^2 x-3\sin x+1=0

for

0∘≤x≤360∘.0^\circ\le x\le 360^\circ.

Step 1: Factor the equation

Treat sin⁡x\sin x like a variable. The expression factors as

2sin⁡2x−3sin⁡x+1=(2sin⁡x−1)(sin⁡x−1).2\sin^2 x-3\sin x+1=(2\sin x-1)(\sin x-1).

Therefore,

(2sin⁡x−1)(sin⁡x−1)=0.(2\sin x-1)(\sin x-1)=0.

Using the zero-product property, either

2sin⁡x−1=02\sin x-1=0

or

sin⁡x−1=0.\sin x-1=0.

So we solve the two equations:

sin⁡x=12\sin x=\frac12

and

sin⁡x=1.\sin x=1.

Step 2: Solve sin⁡x=12\sin x=\frac12

The reference angle is

xr=sin⁡−1(12)=30∘.x_r=\sin^{-1}\left(\frac12\right)=30^\circ.

Sine is positive in Quadrants I and II. Therefore, the angles are

x=30∘x=30^\circ

and

x=180∘−30∘=150∘.x=180^\circ-30^\circ=150^\circ.

Step 3: Solve sin⁡x=1\sin x=1

Sine equals 11 at

x=90∘.x=90^\circ.

Step 4: List the solutions in the interval

All three angles lie between 0∘0^\circ and 360∘360^\circ, so the solutions are

x=30∘, 90∘, 150∘.\boxed{x=30^\circ,\ 90^\circ,\ 150^\circ}.

Step 5: Check

For x=30∘x=30^\circ and x=150∘x=150^\circ, sin⁡x=12\sin x=\frac12:

2(12)2−3(12)+1=12−32+1=0.2\left(\frac12\right)^2-3\left(\frac12\right)+1 =\frac12-\frac32+1=0.

For x=90∘x=90^\circ, sin⁡x=1\sin x=1:

2(1)2−3(1)+1=2−3+1=0.2(1)^2-3(1)+1=2-3+1=0.

All three values satisfy the original equation.

Learn by doing: Solve trigonometric equations over specified intervals

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