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Trigonometric equations

This page is part of the section “Task 1 USE profile level. Elementary equations”. Here are solutions to trigonometric equations with neat table values. Each problem includes the condition, a step-by-step solution, and the answer in general form.

Problems with solutions

Problem 1. Elementary equation with sine

Solve the equation:

sinx=12

1/2 is a table value of sine: sin(π/6) = 1/2. The general formula for the solution of sin x = a:

x=(−1)n·π6+πn,n∈ℤ

Answer: x=(−1)nπ6+πn,n∈ℤ.

Problem 2. Elementary equation with cosine

Solve the equation:

cosx=22

√2/2 is a table value of cosine: cos(π/4) = √2/2. The general formula for the solution of cos x = a:

x=±π4+2πn,n∈ℤ

Answer: x=±π4+2πn,n∈ℤ.

Problem 3. Elementary equation with tangent

Solve the equation:

tgx=1

1 is a table value of tangent: tg(π/4) = 1. The general formula for the solution of tg x = a:

x=π4+πn,n∈ℤ

Answer: x=π4+πn,n∈ℤ.

Problem 4. Quadratic in sin x

Solve the equation:

2sin2x−sinx−1=0

Introduce the substitution t = sin x. Since sine takes values from −1 to 1, t is restricted:

t=sinx,t∈[−1;1]

We obtain a quadratic equation in t:

2t2−t−1=0⇒t1=1,t2=−12

Return to the variable x:

sinx=1⇒x=π2+2πn,n∈ℤ sinx=−12⇒x=(−1)n+1·π6+πn,n∈ℤ

Answer:

x=π2+2πn; x=(−1)n+1π6+πn,n∈ℤ.

Frequently asked questions

How to solve trigonometric equations from Task 1 of the USE?

Elementary equations (sin x = a, cos x = a, tg x = a) are solved by general root formulas. If the equation reduces to a quadratic in sin x or cos x, substitution is used. After substitution it is important to check that the roots fall within the range of the function.

Why does (−1)ⁿ appear in the answer for sin x?

The function sin x takes the same value at two points: x and π − x. The factor (−1)ⁿ allows both families of roots to be written in one formula: for even n we get x = arcsin a + 2πn, for odd n — x = π − arcsin a + 2πn.

Why is the answer for cos x written with ±?

The function cos x is even and takes the same value at x and −x. Therefore both families of roots are written as x = ±arccos a + 2πn.

What to do if after substitution the root does not fall within [−1; 1]?

Such a root is rejected: sin x and cos x cannot take values outside the segment [−1; 1]. For example, if after the substitution t = sin x we get t = 2, the equation sin x = 2 has no solutions.

See also