Answer:
cos3x+tan3x=0
⟹cos3x=−tan3x
⟹cos3x=−sin3xcos3x
⟹cos23x=−sin3x
⟹1−sin23x=−sin3x
⟹sin23x−sin3x−1=0
This is a quadratic equation in sin3x.
sin3x=−(−1)±(−1)2−4×1×(−1)−−−−−−−−−−−−−−−−−√2×1
sin3x=1±5–√2
If x takes real values, the upper sign must be rejected.
sin3x=1−5–√2
⟹3x=nπ+(−1)nsin−11−5–√2
⟹x=13[nπ+(−1)nsin−11−5–√2]
Step-by-step explanation:
Hope this kind of helps
Remember the notation..
When you have a triangle called ABD, the angles may be called in theses equivalent ways>
angle A is equivalent to ∠ DAB
angle B is equivalent to ∠ ABD
angle D is equivalen to ∠ BDA
Then given that the two triangles are congruent, you can say an
angle B = angle F = ∠ EFC
So the answer is the last option of the list.
Use the formula: A=2(3.14)rh+2(3.14)r^2
A=2(3.14)(6)(7)+2(3.14)6^2
A=490.09 so the answer rounded would be 490.1
