B. x=20
Explanation
[tex]\sin (4x-10)=\cos (40-x)[/tex]
by definition
[tex]\sin (\emptyset)=\cos (90-\emptyset)[/tex]
Step 1
let
[tex]\begin{gathered} \emptyset=4x-10 \\ 90-\emptyset=40-x \\ so \\ \sin (4x-10)=\cos (90-(4x-10)) \\ \sin (4x-10)=\cos (90-4x+10) \\ \sin (4x-10)=\cos (100-4x) \\ \text{Also} \\ \sin (4x-10)=\cos (40-x) \\ \text{hence} \\ \cos (40-x)=\cos (100-4x) \\ 40-x=100-4x \\ \text{add x in both sides} \\ 40-x+x=100-4x+x \\ 40=100-3x \\ 40-100=-3x \\ -60=-3x \\ x=\frac{60}{3} \\ x=20 \end{gathered}[/tex]
Let's check every option
A)
[tex]\begin{gathered} \sin (4x-10)=\cos (40-x) \\ x=10 \\ \sin (4\cdot10-10)=\cos (40-10) \\ \sin (30)=\cos (30)\rightarrow false \end{gathered}[/tex]
B)
[tex]\begin{gathered} \sin (4x-10)=\cos (40-x) \\ x=20 \\ \sin (4\cdot20-10)=\cos (40-20) \\ \sin (70)=\cos (20) \\ 70=90-20\rightarrow\text{true} \end{gathered}[/tex]
C)
[tex]\begin{gathered} \sin (4x-10)=\cos (40-x) \\ x=50 \\ \sin (4\cdot50-10)=\cos (40-50) \\ \sin (190)=\cos (-10) \\ 190=90-(-10)\rightarrow\text{false} \end{gathered}[/tex]
D)
[tex]\begin{gathered} \sin (4x-10)=\cos (40-x) \\ x=17 \\ \sin (4\cdot17-10)=\cos (40-17) \\ \sin (58)=\cos (23) \\ 58=90-23\rightarrow\text{false} \end{gathered}[/tex]
therefore, the answer is
B. x=20
