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Basic trigonometric equations

Basic trigonometric equations

The equation sin⁡x=a\displaystyle {\sin x}={a}

The solutions of sin⁡x=a\displaystyle {\sin x}={a} are the numbers whose point on the unit circle is at height a\displaystyle a. On the board, a=32\displaystyle {a}={\frac{\sqrt{3}}{2}}: the line y=32\displaystyle {y}={\frac{\sqrt{3}}{2}} cuts the circle at two points.

  • A=P(π3)\displaystyle {A}={P\left(\frac{\pi}{3}\right)}, and B\displaystyle B is its reflection in the y\displaystyle y-axis: B=P(π−π3)=P(2π3)\displaystyle {B}={P\left(\pi-\frac{\pi}{3}\right)}={P\left(\frac{2\pi}{3}\right)}.
  • Each point belongs to infinitely many numbers, +2πk\displaystyle +2\pi k.
x2+y2=1\displaystyle x^{2}+y^{2}=1
y=32\displaystyle y=\frac{\sqrt{3}}{2}
A=(12,32)\displaystyle A=(\frac{1}{2},\frac{\sqrt{3}}{2})
B=(−12,32)\displaystyle B=(-\frac{1}{2},\frac{\sqrt{3}}{2})

So, when sin⁡α=a\displaystyle {\sin\alpha}={a}:

x=α+2πk,x=π−α+2πk\displaystyle x=\alpha+2\pi k,\quad x=\pi-\alpha+2\pi k

This is the general solution: all the solutions, k\displaystyle k an integer. In the example: x=π3+2πk\displaystyle {x}={\frac{\pi}{3}+2\pi k} or x=2π3+2πk\displaystyle {x}={\frac{2\pi}{3}+2\pi k}.

  • When ∣a∣>1\displaystyle |a|>1 the line does not meet the circle: there is no solution.
  • When a=1\displaystyle {a}={1}, a=−1\displaystyle {a}={-1} or a=0\displaystyle {a}={0} the two families give the same points: sin⁡x=1\displaystyle {\sin x}={1} – x=π2+2πk\displaystyle {x}={\frac{\pi}{2}+2\pi k}; sin⁡x=0\displaystyle {\sin x}={0} – x=πk\displaystyle {x}={\pi k}.
The equation cos⁡x=b\displaystyle {\cos x}={b}

The solutions of cos⁡x=b\displaystyle {\cos x}={b} are the numbers whose point is on the upright line x=b\displaystyle {x}={b}. The two points are reflections of each other in the x\displaystyle x-axis: P(α)\displaystyle P(\alpha) and P(−α)\displaystyle P(-\alpha). When cos⁡α=b\displaystyle {\cos\alpha}={b}:

x=±α+2πk\displaystyle x=\pm\alpha+2\pi k

For example, cos⁡x=−12\displaystyle {\cos x}={-\frac{1}{2}}: cos⁡2π3=−12\displaystyle {\cos\frac{2\pi}{3}}={-\frac{1}{2}}, so x=±2π3+2πk\displaystyle {x}={\pm\frac{2\pi}{3}+2\pi k}. When ∣b∣>1\displaystyle |b|>1 there is no solution.

The equation tan⁡x=c\displaystyle {\tan x}={c}

Every number c\displaystyle c has a point (1,c)\displaystyle (1,c) on the tangent line, and the line from the origin through it cuts the circle at two opposite points – half a turn apart. When tan⁡α=c\displaystyle {\tan\alpha}={c}:

x=α+πk\displaystyle x=\alpha+\pi k

For example, tan⁡x=−1\displaystyle {\tan x}={-1}: x=−π4+πk\displaystyle {x}={-\frac{\pi}{4}+\pi k}. This equation has a solution for every c\displaystyle c.

An α\displaystyle \alpha that is not a special value

When a\displaystyle a is not one of the special values, the calculator (set to radians) gives α\displaystyle \alpha:

  • sin⁡−1a\displaystyle \sin^{-1}a – the number between −π2\displaystyle -\frac{\pi}{2} and π2\displaystyle \frac{\pi}{2} whose sine is a\displaystyle a;
  • cos⁡−1b\displaystyle \cos^{-1}b – the number between 0\displaystyle 0 and π\displaystyle \pi whose cosine is b\displaystyle b;
  • tan⁡−1c\displaystyle \tan^{-1}c – the number between −π2\displaystyle -\frac{\pi}{2} and π2\displaystyle \frac{\pi}{2} whose tangent is c\displaystyle c.

The calculator gives one solution only; the formulas give the rest. For example, sin⁡x=0.3\displaystyle {\sin x}={0.3}:

  • x=sin⁡−10.3+2πk\displaystyle {x}={\sin^{-1}0.3+2\pi k} or x=π−sin⁡−10.3+2πk\displaystyle {x}={\pi-\sin^{-1}0.3+2\pi k};
  • approximately: x≈0.305+2πk\displaystyle x\approx0.305+2\pi k or x≈2.837+2πk\displaystyle x\approx2.837+2\pi k.
The solutions in a given interval

To find the solutions in a given interval, write the general solution and substitute values of k\displaystyle k until you leave the interval.

The example below solves cos⁡x=−12\displaystyle {\cos x}={-\frac{1}{2}} in the interval 0≤x≤2π\displaystyle 0\leq x\leq2\pi.

  • The general solution: x=2π3+2πk\displaystyle {x}={\frac{2\pi}{3}+2\pi k} or x=−2π3+2πk\displaystyle {x}={-\frac{2\pi}{3}+2\pi k}.
  • In the first family, k=0\displaystyle {k}={0} gives 2π3\displaystyle \frac{2\pi}{3}; k=1\displaystyle {k}={1} is already greater than 2π\displaystyle 2\pi.
  • In the second family, k=0\displaystyle {k}={0} gives a negative number; k=1\displaystyle {k}={1} gives 4π3\displaystyle \frac{4\pi}{3}.

The solutions: x=2π3\displaystyle {x}={\frac{2\pi}{3}} or x=4π3\displaystyle {x}={\frac{4\pi}{3}}.

Exercise 1

Solve. Write the general solution.

a
    b
      c
        d
          e
            Exercise 2

            Solve with sin⁡−1\displaystyle \sin^{-1} and tan⁡−1\displaystyle \tan^{-1}. Write the general solution.

            a
              b
                Exercise 3

                Solve in the given interval.

                a

                0≤x≤2π\displaystyle 0\leq x\leq2\pi

                  b

                  −π≤x≤π\displaystyle -\pi\leq x\leq\pi

                    c

                    0≤x≤4π\displaystyle 0\leq x\leq4\pi