You cannot cancel 2 cos x away.

Wait for my working

ok! Thx

7 sin x cos² x = 2 cos x

7 sin x cos² x - 2 cos x = 0

(7 sin x cos x - 2)(cos x) = 0

7 sin x cos x = 2 or cos x = 0

7/2 sin 2x = 2

(Recall that sin 2x = 2 sin x cos x)

sin 2x = 2 × 2/7 = 4/7


For cos x = 0,

angle = π/2 rad, 3π/2 rad, ...
But due to the range, only these two fit.


For sin 2x = 4/7,

Basic angle = sin-¹ (4/7) ≈ 0.60825 rad

4/7 is positive so we look at the 1st and 2nd quadrant

Since 0 ≤ x ≤ 2π, then 0 ≤ 2x ≤ 4π


2x ≈ 0.60825 rad , (π - 0.60825) rad , (2π + 0.60825) rad , (3 π - 0.60825) rad

2x ≈ 0.60825 rad, 2.5334 rad, 6.8914 rad, 8.8165 rad

x ≈ 0.30413 rad, 1.2667 rad, 3.4457 rad, 4.4083 rad

x = 0.304 rad, 1.27 rad, 3.45 rad, 4.41 rad
(To 3s.f)

Main takeaway : don't cancel terms as you will lose a possible solution.

Some analogies :


x² = 2x

We cannot just divide both sides by x. We would lose one possible solution (x = 0)

Bring over and factorise.

x² - 2x = 0

x(x - 2) = 0

x = 0 or x = 2


Another one :

7yx² = 2x

Same thing here.

7yx² - 2x = 0

x(7xy - 2) = 0

x = 0

or

7xy = 2

xy = 2/7

x = 2/7y

(or in terms of y, y = 2/7x)

Thank you so much! I got it

Thx for the analogy as well!!!!

Welcome