Worked examples · Sheet 3

Quadratic equations in sine or cosine

Theory used: Quadratic equations in sine or cosine Pythagorean identity Elementary equations

Exercise 1

Solve the equation

2 sin^{2} x - 3 sin x + 1 = 0.

With the substitution $t = sin x$ one obtains the system

{2 t^{2} - 3 t + 1 = 0 - 1 \leq t \leq 1

The equation in $t$ has discriminant $Δ = 9 - 8 = 1$ and roots

t = \frac{3 \pm 1}{4}, t_{1} = 1, t_{2} = \frac{1}{2},

both acceptable. From $sin x = 1$ :

x = \frac{π}{2} + 2 nπ .

From $sin x = \frac{1}{2}$ :

x = \frac{π}{6} + 2 nπ or x = \frac{5 π}{6} + 2 nπ .

Solutions: $x = \frac{π}{2} + 2 nπ$ , $x = \frac{π}{6} + 2 nπ$ or $x = \frac{5 π}{6} + 2 nπ$ , with $n \in Z$ .

The equation contains only $sin x$ and its square, so the substitution $t = sin x$ turns it into an ordinary quadratic equation, together with the condition $- 1 \leq t \leq 1$ because the sine takes only values in that interval. Both roots satisfy the condition, hence each gives an elementary equation. The value $sin x = 1$ identifies a single point on the unit circle, the highest point, therefore it produces a single family; the value $sin x = \frac{1}{2}$ identifies two points, of angles $\frac{π}{6}$ and $\frac{5 π}{6}$ , therefore it produces two families.

Exercise 2

Solve the equation

2 cos^{2} x + 5 cos x + 2 = 0.

With the substitution $t = cos x$ one obtains the system

{2 t^{2} + 5 t + 2 = 0 - 1 \leq t \leq 1

The discriminant is $Δ = 25 - 16 = 9$ and the roots are

t = \frac{- 5 \pm 3}{4}, t_{1} = - \frac{1}{2}, t_{2} = - 2.

The root $t_{2} = - 2$ does not satisfy the condition $- 1 \leq t \leq 1$ and is discarded. From $cos x = - \frac{1}{2}$ :

x = \pm \frac{2 π}{3} + 2 nπ .

Solutions: $x = \pm \frac{2 π}{3} + 2 nπ$ , with $n \in Z$ .

The procedure is the same as in the previous exercise, with the cosine in place of the sine. Here one of the two roots of the equation in $t$ equals $- 2$ , and since the cosine takes only values between $- 1$ and $1$ , the equation $cos x = - 2$ has no solutions: that root does not contribute. What remains is $cos x = - \frac{1}{2}$ , an elementary equation whose solutions are the angles $\pm \frac{2 π}{3}$ up to full turns, symmetric with respect to the horizontal axis.

Exercise 3

Solve the equation

2 sin^{2} x + 3 cos x - 3 = 0.

The equation contains both $sin^{2} x$ and $cos x$ , so one applies the Pythagorean identity in the form $sin^{2} x = 1 - cos^{2} x$ :

2 (1 - cos^{2} x) + 3 cos x - 3 - 2 cos^{2} x + 3 cos x - 1 2 cos^{2} x - 3 cos x + 1 = 0 = 0 = 0.

With $t = cos x$ : the discriminant is $Δ = 9 - 8 = 1$ and the roots are

t = \frac{3 \pm 1}{4}, t_{1} = 1, t_{2} = \frac{1}{2},

both acceptable. From $cos x = 1$ :

x = 2 nπ .

From $cos x = \frac{1}{2}$ :

x = \pm \frac{π}{3} + 2 nπ .

Solutions: $x = 2 nπ$ or $x = \pm \frac{π}{3} + 2 nπ$ , with $n \in Z$ .

The direct substitution does not apply right away, because the equation contains two different functions. The square of the sine, however, can be rewritten with the Pythagorean identity, and after the substitution the whole equation contains only the cosine; the final change of sign serves to make the coefficient of the quadratic term positive. From here the procedure is the standard one: quadratic equation in $t$ , check of the condition $- 1 \leq t \leq 1$ , then an elementary equation for each acceptable root. A check: for $x = \frac{π}{3}$ one has $2 \cdot \frac{3}{4} + 3 \cdot \frac{1}{2} - 3 = \frac{3}{2} + \frac{3}{2} - 3 = 0$ , therefore the equation is satisfied.