First step – find θ

 tan 2θ = 3

2θ = arctan 3

2θ = 71.57° (to 2 decimal place)

θ = 35.8°

ps: arc tan is the same as inverse tan

pss: avoid the common error of doing divide by 2 first, then doing arctan (1.5), also check you are working in degrees and not radians in your calculator.

Second step – find the other θ values between 0°< θ < 360°

This part was the tricky part.

Let us start from the beginning tan θ graph looks as follows:

As you can see there is an asymptote after every 180°, right? Between -90 and 90° is a gap of 180°.

If tan θ = 0 (y=0 line, which is the x-axis), then the next time the graph crosses the x axis is at 180°, 360° and so on.

We can write this as θ + 180n (for any integer n), right?

Another way to look at this is through the dreaded but impressive quadrants!

As you can see tan θ = tan (180°+θ )

Let’s find out the values then but wait!

Let’s go back to your GCSE days and think about graph transformation because remember the question isn’t tan θ =3, but actually tan 2θ =3.

Since the transformation is inside the bracket, it affects the x axis and so we do the opposite. Instead of multiplying by 2, we divide the x points by 2.

tan θ  = 3

tan 2θ  = 3

The answers

As you can see for tan 2θ  there are 4 values between the range of 0 and 360°, and that is because the asymptotes have changed. Since all the angles are being divided by 2, the asymptotes are now repeating every 90°.

Your answers for θ therefore are θ+90n (for any integer n)

35.8°, 125.8°, 215.8° and 305.8°

I hope this helps.

Your Turn

Solve the equation cos 2θ = 0.3 for 0° ≤ θ < 360°. Use the comments section below to write your answer.