Solve Trigonometric Equations - Grade 12

\(\sin(x)\) is positive in quadrant I and II and therefore the given equation has two solutions.

In quadrant I, the solution to \(\sin(x) = \frac{1}{2}\) is \[x = \frac{\pi}{6}\] (green angle in unit circle).

In quadrant II, and by symmetry of the unit circle, the solution to \(\sin(x) = \frac{1}{2}\) is \[ x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \] (brown angle in unit circle).

The graphical solutions of the equation \(\sin(x) = \frac{1}{2}\) are obtained by first writing the equation with right side equal to zero: \(\sin(x) - \frac{1}{2} = 0\). Then, graph the left side of the equation, and the solutions of the equation are the x-intercepts of the graph of \(f(x) = \sin(x) - \frac{1}{2}\), as shown below. It is easy to check that the two x-intercepts in the graph below are close to the analytical solutions found above: \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\).

The equation may be written as: \[ \sin(x) = -\cos(x) \] Divide both sides of the equation by \(\cos(x)\): \[ \frac{\sin(x)}{\cos(x)} = -1 \] Use the identity \(\tan(x) = \frac{\sin(x)}{\cos(x)}\) to rewrite the equation as: \[ \tan(x) = -1 \] Solve the equation \(\tan(x) = -1\) \(\tan(x)\) is negative in quadrant II and IV. Solution in quadrant II: \[ x = \frac{3\pi}{4} \] By symmetry of the unit circle, the solution in quadrant IV is given by: \[ \frac{3\pi}{4} + \pi = \frac{7\pi}{4} \]

The graphical solutions may be approximated by the x-intercepts of the graph of \( g(x) = \sin(x) + \cos(x) \). We can easily check that the two solutions found above, \(\frac{3\pi}{4}\) and \(\frac{7\pi}{4}\), are close to the x-intercepts of the graph shown below.

We first use the formula \[ \sin(A - B) = \sin A \cos B - \cos A \sin B \] to write: \[ \sin\left(\frac{\pi x}{2}\right)\cos\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi x}{2}\right)\sin\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi x}{2} - \frac{\pi}{3}\right) \] Rewrite the given equation as follows: \[ \sin\left(\frac{\pi x}{2} - \frac{\pi}{3}\right) = 0 \] Solve the above equation to obtain: \[ \frac{\pi x}{2} - \frac{\pi}{3} = k\pi \quad \text{where} \quad k = 0, \pm1, \pm2, \ldots \] Simplify and rewrite the solution as: \[ x = 2k + \frac{2}{3} \quad \text{where} \quad k = 0, \pm1, \pm2, \ldots \]

The first 3 positive solutions are: \[ \frac{2}{3}, \quad \frac{8}{3}, \quad \frac{14}{3} \] corresponding to \( k = 0, 1, 2 \). These values are close to the x-intercepts of the graph of \[ f(x) = \sin\left(\frac{\pi x}{2}\right)\cos\left(\frac{\pi}{3}\right) - \cos\left(\frac{\pi x}{2}\right)\sin\left(\frac{\pi}{3}\right) \] shown above.

We first use the identity: \[ \sin^2(x) = 1 - \cos^2(x) \] to rewrite the equation as: \[ 2(1 - \cos^2(x)) + \cos(x) = 1 \] Group like terms and write the equation with the right side equal to zero: \[ -2 \cos^2(x) + \cos(x) + 1 = 0 \] Let \( u = \cos(x) \) and rewrite the equation as: \[ -2u^2 + u + 1 = 0 \] Solve for \( u \) to obtain: \( \quad u = 1 \) and \( \quad u = -\frac{1}{2} \)

1) Solve \( u = 1 \) for \( x \): \[ u = \cos(x) = 1 \] to obtain \[ x_1 = 2k\pi \quad \text{where } k = 0, \pm 1, \pm 2, \ldots \] 2) Solve \( u = -\frac{1}{2} \) for \( x \): \[ \cos(x) = -\frac{1}{2} \] \( \cos(x) \) is negative in quadrants II and III, and therefore \( \cos(x) = -\frac{1}{2} \) has two more groups of solutions. \[ x_2 = \frac{2\pi}{3} + 2k\pi \quad \text{where } k = 0, \pm 1, \pm 2, \ldots \] and \[ x_3 = \frac{4\pi}{3} + 2k\pi \quad \text{where } k = 0, \pm 1, \pm 2, \ldots \] The first 3 positive solutions are obtained by setting \( k = 0 \) in all three groups of solutions to obtain: \[ 0,\quad \frac{2\pi}{3},\quad \frac{4\pi}{3} \] These values are close to the \( x \)-intercepts of the graph of \( g(x) = 2\sin^2(x) + \cos(x) - 1 \) shown below.

We first use the identity: \[ \cos(x) = 2\cos^2\left(\frac{x}{2}\right) - 1 \] to rewrite the equation as: \[ 6\cos^2\left(\frac{x}{2}\right) - \left(2\cos^2\left(\frac{x}{2}\right) - 1\right) = 4 \] Group like terms and write the equation with the right side equal to zero: \[ 4\cos^2\left(\frac{x}{2}\right) = 3 \] \[ \cos\left(\frac{x}{2}\right) = \pm \frac{\sqrt{3}}{2} \] 1) Solve the first equation \(\cos\left(\frac{x}{2}\right) = \frac{\sqrt{3}}{2}\) The cosine function is positive in quadrant I and IV hence the two groups of solutions: \[ \frac{x}{2} = \frac{\pi}{6} + 2k\pi \quad \text{where } k = 0, \pm1, \pm2, \ldots \quad \text{(quadrant I)} \] and \[ \frac{x}{2} = \frac{11\pi}{6} + 2k\pi \quad \text{where } k = 0, \pm1, \pm2, \ldots \quad \text{(quadrant IV)} \] Multiply all terms by 2 to obtain the first two groups of solutions: \[ x_1 = \frac{\pi}{3} + 4k\pi \quad \text{where } k = 0, \pm1, \pm2, \ldots \] \[ x_2 = \frac{11\pi}{3} + 4k\pi \quad \text{where } k = 0, \pm1, \pm2, \ldots \] 2) Solve the second equation \(\cos\left(\frac{x}{2}\right) = -\frac{\sqrt{3}}{2}\) The cosine function is negative in quadrant II and III hence the two groups of solutions: \[ \frac{x}{2} = \frac{5\pi}{6} + 2k\pi \quad \text{where } k = 0, \pm1, \pm2, \ldots \quad \text{(quadrant II)} \] and \[ \frac{x}{2} = \frac{7\pi}{6} + 2k\pi \quad \text{where } k = 0, \pm1, \pm2, \ldots \quad \text{(quadrant III)} \] Multiply all terms by 2 to obtain two more groups of solutions: \[ x_3 = \frac{5\pi}{3} + 4k\pi \quad \text{where } k = 0, \pm1, \pm2, \ldots \] \[ x_4 = \frac{7\pi}{3} + 4k\pi \quad \text{where } k = 0, \pm1, \pm2, \ldots \] The first 4 positive solutions are obtained by setting \(k = 0\) in all four groups of solutions to obtain: \(\frac{\pi}{3}, \frac{5\pi}{3}, \frac{7\pi}{3}, \frac{11\pi}{3}\). These values are close to the x-intercepts of the graph of \[ f(x) = 6\cos^2\left(\frac{x}{2}\right) - \cos(x) - 4 \] shown below.

We first use the identity: \[ \tan^{2}(x) = \sec^{2}(x) - 1 \] to rewrite the equation as: \[ \sec^{2}(x) - 1 + 2 \sec(x) + 1 = 0 \] Group like terms and write the equation with the right side equal to zero: \[ \sec^{2}(x) + 2 \sec(x) = 0 \] Factor: \[ \sec(x) \bigl(\sec(x) + 2\bigr) = 0 \] \(\sec(x) = 0\) has no solution.

Solve: \[ \sec(x) + 2 = 0 \] Use \(\sec(x) = \frac{1}{\cos(x)}\) to rewrite the equation as: \[ \cos(x) = -\frac{1}{2} \] The cosine function is negative in quadrant II and III; hence two groups of solutions: \[ x_1 = \frac{2\pi}{3} + 2k\pi \quad \text{where} \quad k = 0, \pm 1, \pm 2, \ldots \] \[ x_2 = \frac{4\pi}{3} + 2k\pi \quad \text{where} \quad k = 0, \pm 1, \pm 2, \ldots \]

The first 2 positive solutions are obtained by setting \(k = 0\) in the two groups of solutions to obtain: \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\). These values are close to the x intercepts of the graph of \(f(x) = \tan^{2}(x) + 2 \sec(x) + 1\) shown above.

We first use the identities: \[ \cos(3x) = \cos^{3}(x) - 3 \cos(x) \sin^{2}(x) \] and \[ \sin(2x) = 2 \sin(x) \cos(x) \] to rewrite the given equation as follows: \[ \cos^{3}(x) - 3 \cos(x) \sin^{2}(x) + 2 \sin(x) \cos(x) = \cos(x) \] Rewrite with right hand side equal to zero and factor \(\cos(x)\): \[ \cos(x) \left( \cos^{2}(x) - 3 \sin^{2}(x) + 2 \sin(x) - 1 \right) = 0 \] Use the identity \(\cos^{2}(x) = 1 - \sin^{2}(x)\) to rewrite equation as follows: \[ \cos(x) \left( 1 - \sin^{2}(x) - 3 \sin^{2}(x) + 2 \sin(x) - 1 \right) = 0 \] Simplify: \[ \cos(x) \left( -4 \sin^{2}(x) + 2 \sin(x) \right) = 0 \] Factor: \[ 2 \sin(x) \cos(x) (1 - 2 \sin(x)) = 0 \] Set each factor equal to zero and solve. \[ \sin(x) = 0 \quad \Rightarrow \quad x_1 = k \pi \] \[ \cos(x) = 0 \quad \Rightarrow \quad x_2 = \frac{\pi}{2} + k \pi \] \[ 1 - 2 \sin(x) = 0 \quad \text{or} \quad \sin(x) = \frac{1}{2} \] gives two other groups of solutions: \[ x_3 = \frac{\pi}{6} + 2 k \pi \] and \[ x_4 = \frac{5 \pi}{6} + 2 k \pi \]

The first 6 positive solutions are obtained by setting \(k = 0\) and \(k = 1\) in the first two groups of solutions \(x_1\) and \(x_2\) to obtain: \[ 0, \quad \pi, \quad \frac{\pi}{2}, \quad \frac{3 \pi}{2} \] and \(k=0\) in the second and third groups of solutions \(x_3\) and \(x_4\) to obtain \[ \frac{\pi}{6} \quad \text{and} \quad \frac{5 \pi}{6} \] These 6 solutions are close to the \(x\)-intercepts of the graph of \[ f(x) = \cos(3x) + \sin(2x) - \cos(x) \] shown below.

We use the fact that if \(\cos(A) = \cos(B)\), then \[ A = B + 2k\pi \quad \text{or} \quad A = -B + 2k\pi \] to write: \[ 3x = 2x + \frac{\pi}{4} + 2k\pi \] or \[ 3x = -(2x + \frac{\pi}{4}) + 2k\pi \] Solve \[ 3x = 2x + \frac{\pi}{4} + 2k\pi \] Solutions: \[ x_1 = \frac{\pi}{4} + 2k\pi \] Solve \[ 3x = -(2x + \frac{\pi}{4}) + 2k\pi \] gives \[ 5x = -\frac{\pi}{4} + 2k\pi \] Divide all terms by 5: \[ x_2 = -\frac{\pi}{20} + \frac{2k\pi}{5} \] where \( k = 0, \pm 1, \pm 2, \ldots \) in both groups of solutions.

The 6 x-intercepts shown in the graph of \[ f(x) = \cos(3x) - \cos\left(2x + \frac{\pi}{4}\right) \] shown below correspond to the solutions \(k = 0\) in the first group \(x_1\) and \(k = 0, 1, 2, 3, 4\) in the second group of solutions \[ x_2 = -\frac{\pi}{20} + \frac{2k\pi}{5} . \]