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

Sagot :

To find the value of [tex]\( x \)[/tex] given the equation [tex]\(\cos x = \sin (20 + x)^{\circ}\)[/tex] for [tex]\(0^{\circ} < x < 90^{\circ}\)[/tex]:

1. Recall the co-function identity for trigonometric functions: [tex]\(\cos \theta = \sin (90^{\circ} - \theta)\)[/tex].
2. Use this identity to rewrite the given equation in terms of a sine function:
[tex]\[ \cos x = \sin (90^{\circ} - x) \][/tex]
3. Therefore, we can equate:
[tex]\[ \sin (90^{\circ} - x) = \sin (20^{\circ} + x) \][/tex]
4. Since the sine function is periodic, the general solution of [tex]\(\sin A = \sin B\)[/tex] is given by:
[tex]\[ 90^{\circ} - x = 20^{\circ} + x \][/tex]
(Or the other general solutions involving [tex]\(\pi\)[/tex] periods, which are not relevant here because we work in degrees between 0 and 90).
5. Solve the equation:
[tex]\[ 90^{\circ} - x = 20^{\circ} + x \][/tex]
6. Combine like terms:
[tex]\[ 90^{\circ} - 20^{\circ} = x + x \][/tex]
7. Simplify:
[tex]\[ 70^{\circ} = 2x \][/tex]
8. Divide both sides by 2:
[tex]\[ x = \frac{70}{2} = 35^{\circ} \][/tex]

So, the value of [tex]\( x \)[/tex] is [tex]\( \boxed{35} \)[/tex].