Further Trigonometric Equations0%

Trigonometry · Topic 12 of 19

Further Trigonometric Equations

Video lesson · from 45:463 worked examples

One lesson video covers all of Trigonometry, so it opens at 45:46 for this topic — not from the beginning.

Theory

Trigonometric equations which contain a mixture of double angles (like 2x2x) and single angles (like xx) require using Double Angle Formulae to find solutions.

The goal is to substitute the double angle term so that the entire equation is in terms of the same single angle trig function, which often results in a quadratic equation you can factorise.

⚠️ Common Examiner Traps

  • Never divide by a trig term — factorise it out. Dividing an equation by cos⁡x\cos x loses solutions. Dividing throws away every solution where cos⁡x=0\cos x = 0. Take the common factor out and set each factor to zero instead.
  • Extract the common factor: many candidates do not spot one at all. After substituting the double angle formula, always look for a shared sin⁡x\sin x or cos⁡x\cos x.
  • Pick the right form of cos⁡2x\cos 2x: choose whichever of the three versions leaves the equation in one trig function only. Match it to the other term in the equation.
  • Give every solution in the domain: each factor produces its own set of answers. Use a quadrant diagram and check you have them all before you stop.

Worked examples

Example 1

Solve sin⁡2x∘+sin⁡x∘=0\sin 2x^\circ + \sin x^\circ = 0 for 0≤x∘<3600 \leq x^\circ < 360.

Replace sin⁡2x∘\sin 2x^\circ using the exact formula 2sin⁡x∘cos⁡x∘2\sin x^\circ\cos x^\circ:

2sin⁡x∘cos⁡x∘+sin⁡x∘=02\sin x^\circ\cos x^\circ + \sin x^\circ = 0

Factorise by taking out the common factor of sin⁡x∘\sin x^\circ:

sin⁡x∘(2cos⁡x∘+1)=0\sin x^\circ(2\cos x^\circ + 1) = 0

This gives two equations to solve:

sin⁡x∘=0\sin x^\circ = 0

From the sine graph:

x=0,180x = 0, 180

Note: 360 is not in the domain (<360<360).

2cos⁡x∘+1=0cos⁡x∘=−12\begin{aligned} 2\cos x^\circ + 1 &= 0 \\ \cos x^\circ &= -\frac{1}{2} \end{aligned}

Base angle: 60∘60^\circ. Q2, Q3.

x=180−60=120x=180+60=240\begin{aligned} x &= 180 - 60 = 120 \\ x &= 180 + 60 = 240 \end{aligned}

Solutions: 0°, 120°, 180°, 240°

Example 2

Solve cos⁡2x=cos⁡x\cos 2x = \cos x for 0≤x≤2π0 \leq x \leq 2\pi.

For cos⁡2x\cos 2x, we have three choices. Since the other term is cos⁡x\cos x, we should pick the formula that only contains cosine: 2cos⁡2x−12\cos^2 x - 1.

2cos⁡2x−1=cos⁡x2cos⁡2x−cos⁡x−1=0\begin{aligned} 2\cos^2 x - 1 &= \cos x \\ 2\cos^2 x - \cos x - 1 &= 0 \end{aligned}

Factorise (let u=cos⁡xu = \cos x):

(2cos⁡x+1)(cos⁡x−1)=0(2\cos x + 1)(\cos x - 1) = 0
2cos⁡x+1=0cos⁡x=−12\begin{aligned} 2\cos x + 1 &= 0 \\ \cos x &= -\frac{1}{2} \end{aligned}

Base angle: π3\frac{\pi}{3}. Q2, Q3.

x=π−π3=2π3x=π+π3=4π3\begin{aligned} x &= \pi - \frac{\pi}{3} = \frac{2\pi}{3} \\ x &= \pi + \frac{\pi}{3} = \frac{4\pi}{3} \end{aligned}
cos⁡x=1\cos x = 1

From the cosine graph:

x=0,2πx = 0, 2\pi

Solutions: 00, 2π3\frac{2\pi}{3}, 4π3\frac{4\pi}{3}, 2π2\pi

Example 3

Find the points of intersection of the graphs y=3sin⁡2x+1y = 3\sin 2x + 1 and y=2sin⁡x+1y = 2\sin x + 1 for 0≤x≤2π0 \leq x \leq 2\pi.

Set them equal to find the intersection points:

3sin⁡2x+1=2sin⁡x+13sin⁡2x=2sin⁡x\begin{aligned} 3\sin 2x + 1 &= 2\sin x + 1 \\ 3\sin 2x &= 2\sin x \end{aligned}

Substitute sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x:

3(2sin⁡xcos⁡x)=2sin⁡x6sin⁡xcos⁡x−2sin⁡x=0\begin{aligned} 3(2\sin x\cos x) &= 2\sin x \\ 6\sin x\cos x - 2\sin x &= 0 \end{aligned}

Factorise by taking out 2sin⁡x2\sin x:

2sin⁡x(3cos⁡x−1)=02\sin x(3\cos x - 1) = 0
2sin⁡x=0sin⁡x=0\begin{aligned} 2\sin x &= 0 \\ \sin x &= 0 \end{aligned}

From the sine graph:

x=0,π,2πx = 0, \pi, 2\pi
3cos⁡x−1=0cos⁡x=13\begin{aligned} 3\cos x - 1 &= 0 \\ \cos x &= \frac{1}{3} \end{aligned}

Base angle ≈1.231\approx 1.231 rad. Q1, Q4.

x=1.231x=2π−1.231=5.052\begin{aligned} x &= 1.231 \\ x &= 2\pi - 1.231 = 5.052 \end{aligned}

Usually, "points of intersection" requires coordinates. Substitute these x-values back into either original equation (e.g., y=2sin⁡x+1y = 2\sin x + 1) to find y-coordinates.

  • When x=0x = 0, y=2(0)+1=1y = 2(0) + 1 = 1. Point: (0,1)(0, 1)
  • When x=πx = \pi, y=2(0)+1=1y = 2(0) + 1 = 1. Point: (π,1)(\pi, 1)
  • When x=2πx = 2\pi, y=2(0)+1=1y = 2(0) + 1 = 1. Point: (2π,1)(2\pi, 1)
  • When x=1.231x = 1.231, sin⁡(1.231)=1−(1/3)2=83\sin(1.231) = \sqrt{1 - (1/3)^2} = \frac{\sqrt{8}}{3}. y=2(83)+1=28+33≈2.89y = 2(\frac{\sqrt{8}}{3}) + 1 = \frac{2\sqrt{8}+3}{3} \approx 2.89. Point: (1.231,2.89)(1.231, 2.89)
  • When x=5.052x = 5.052, sin⁡(5.052)=−83\sin(5.052) = -\frac{\sqrt{8}}{3}. y=2(−83)+1=3−283≈−0.89y = 2(-\frac{\sqrt{8}}{3}) + 1 = \frac{3-2\sqrt{8}}{3} \approx -0.89. Point: (5.052,−0.89)(5.052, -0.89)