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If [tex]\(\cos x = \sin (20^\circ + x)\)[/tex] and [tex]\(0^\circ \ \textless \ x \ \textless \ 90^\circ\)[/tex], the value of [tex]\(x\)[/tex] is [tex]\(\square^\circ\)[/tex].

Sagot :

To find the value of [tex]\( x \)[/tex] that satisfies the equation [tex]\(\cos x = \sin (20^\circ + x)\)[/tex], follow these detailed steps:

1. Understand the trigonometric identities: One key identity to remember is that [tex]\(\cos \theta = \sin (90^\circ - \theta)\)[/tex]. This means we can rewrite [tex]\(\cos x\)[/tex] as [tex]\(\sin (90^\circ - x)\)[/tex].

2. Rewrite the equation: Using the identity, we can express [tex]\(\cos x\)[/tex] as [tex]\(\sin (90^\circ - x)\)[/tex]. So the original equation becomes:
[tex]\[ \sin (90^\circ - x) = \sin (20^\circ + x) \][/tex]

3. Set the arguments equal: Since the sine function has the property that [tex]\(\sin A = \sin B\)[/tex] implies [tex]\(A = B + 360^\circ k\)[/tex] or [tex]\(A = 180^\circ - B + 360^\circ k\)[/tex] for any integer [tex]\(k\)[/tex], apply this property. However, given that [tex]\(0^\circ < x < 90^\circ\)[/tex], we can directly set the angles equal without considering multiple cycles:
[tex]\[ 90^\circ - x = 20^\circ + x \][/tex]

4. Solve for [tex]\( x \)[/tex]: Combine like terms to isolate [tex]\( x \)[/tex].
[tex]\[ 90^\circ - x = 20^\circ + x \][/tex]
Add [tex]\( x \)[/tex] to both sides:
[tex]\[ 90^\circ = 20^\circ + 2x \][/tex]
Subtract [tex]\( 20^\circ \)[/tex] from both sides:
[tex]\[ 70^\circ = 2x \][/tex]
Divide by 2:
[tex]\[ x = 35^\circ \][/tex]

So, the value of [tex]\( x \)[/tex] is [tex]\( 35^\circ \)[/tex].

Therefore, if [tex]\(\cos x = \sin (20^\circ + x)\)[/tex] and [tex]\(0^\circ < x < 90^\circ\)[/tex], the value of [tex]\( x \)[/tex] is [tex]\( \boxed{35^\circ} \)[/tex].