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Solving second degree trigonometric equations

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Solving Second-Degree Trigonometric Equations

A second-degree trigonometric equation has the trig function squared, structured like a quadratic. Substitute a variable such as y equals sine x to turn it into a standard quadratic, factor or use the quadratic formula, then solve each resulting first-degree trig equation, with a full worked example.

What a second-degree trig equation is

A second-degree trigonometric equation has the trig function squared, such as 2sin²x − sin x − 1 = 0. It's structured exactly like a quadratic equation, just with sin x (or cos x, tan x) standing in for the variable.

Solving a second-degree trig equation Solve 2 sine squared x minus sine x minus 1 equals 0 for x from 0 to 360 degrees. Substitute y for sin x to get 2y squared minus y minus 1 equals 0, factor to (2y+1)(y-1)=0, giving sin x equals 1 or sin x equals negative one half. Solve: 2sin²x − sin x − 1 = 0, for 0° ≤ x < 360° Step 1Substitute y = sin x: 2y² − y − 1 = 0 Step 2Factor: (2y + 1)(y − 1) = 0 Step 3So y = −1/2 or y = 1, meaning sin x = −1/2 or sin x = 1 Step 4sin x = 1 gives x = 90° Step 5sin x = −1/2 gives x = 210° and x = 330°
Substituting y for the trig ratio turns the equation into an ordinary quadratic you can factor.

The substitution method

The cleanest approach is to substitute a variable, such as y = sin x, which turns the trig equation into a standard quadratic: 2y² − y − 1 = 0. This quadratic can be factored or solved with the quadratic formula just like any other.

Worked example

Solve 2sin²x − sin x − 1 = 0 for 0° ≤ x < 360°. Substituting y = sin x gives 2y² − y − 1 = 0, which factors to (2y + 1)(y − 1) = 0. So y = −1/2 or y = 1, meaning sin x = −1/2 or sin x = 1.

Solving sin x = 1 gives x = 90°. Solving sin x = −1/2 (sine negative, in quadrants III and IV, reference angle 30°) gives x = 210° and x = 330°. The complete solution set is x = 90°, 210°, 330°.

Why this connects to first-degree equations

Once the quadratic is factored, each factor becomes its own first-degree trig equation to solve using the reference-angle-and-quadrant method.

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