Find the value of x, in degrees, for 0^\circ \le x \le 90^\circ that satisfies the trigono… — JAMB UTME

Find the value of $x$ , in degrees, for $0^\circ \le x \le 90^\circ$ that satisfies the trigonometric equation $\sin 2x = \cos(x + 30^\circ)$ .

Answer: 20 degrees

The value of

x

in the interval

0^\circ \le x \le 90^\circ

satisfying the equation is

20^\circ

Using the co-function identity

\cos \alpha = \sin(90^\circ - \alpha)

, we convert the right-hand side to

\sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x)

. Equating the arguments gives

2x = 60^\circ - x

, which simplifies to

3x = 60^\circ

, yielding

x = 20^\circ

Apply the co-function trigonometric identity

\cos(x + 30^\circ) = \sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x)

Converting cosine to sine allows direct comparison of sine functions on both sides of the equation.

Set up the equation equating the angle expressions

2x = 60^\circ - x

Since

\sin(2x) = \sin(60^\circ - x)

and

x \in [0^\circ, 90^\circ]

, equating the principal angle arguments gives the primary solution.

Solve the linear equation for

x

3x = 60^\circ \implies x = 20^\circ

Adding

x

to both sides gives

3x = 60^\circ

, and dividing by 3 yields

x = 20^\circ

Co-function identities and simple trigonometric equations

Question