At which values of x within the domain 0^\circ \le x \le 360^\circ do the graphs of f(x) =… — JAMB UTME

At which values of $x$ within the domain $0^\circ \le x \le 360^\circ$ do the graphs of $f(x) = 2\sin x$ and $g(x) = \tan x$ intersect?

The graphs intersect at

x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

Equating

2\sin x = \frac{\sin x}{\cos x}

yields

\sin x (2\cos x - 1) = 0

. Setting

\sin x = 0

gives

x = 0^\circ, 180^\circ, 360^\circ

, and setting

\cos x = \frac{1}{2}

gives

x = 60^\circ, 300^\circ

. Combining these gives the full solution set.

Set the two trigonometric expressions equal to each other to find points of intersection.

2\sin x = \tan x

Intersection points occur where

f(x) = g(x)

Rewrite

\tan x

in terms of sine and cosine.

2\sin x = \frac{\sin x}{\cos x}

, for

x \neq 90^\circ, 270^\circ

Using the quotient identity

\tan x = \frac{\sin x}{\cos x}

allows simplification.

Rearrange the equation and factor out

\sin x

2\sin x \cos x - \sin x = 0 \implies \sin x(2\cos x - 1) = 0

Factoring prevents losing solutions that occur when

\sin x = 0

Solve each factor separately within

0^\circ \le x \le 360^\circ

Factor 1:

\sin x = 0 \implies x = 0^\circ, 180^\circ, 360^\circ

.
Factor 2:

2\cos x - 1 = 0 \implies \cos x = \frac{1}{2} \implies x = 60^\circ, 300^\circ

Cosine is positive in the first and fourth quadrants.

Combine all valid solutions.

x = 0^\circ, 60^\circ, 180^\circ, 300^\circ, 360^\circ

All five values satisfy the original equation and lie within the given domain.

Solving trigonometric equations by factoring and finding all roots in a given domain

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