Question

What are the degree measures of all positive angles between 0 and 90 which satisfy the equation: \[\sin^2x + \cos^2x+\tan^2x+\cot^2x+\sec^2x+\csc^2x=31\]

Answer 1

really? i would start with
\[\tan^2(x)+\csc^2(x)=15\] but then i would be stuck

Answer 2

so it's not possible?

Answer 3

is possible, PITA tho

Answer 4

Rewrite:
\[\sin ^{2}(x)+\cos ^{2}(x)=1;\sec ^{2}(x)=\tan ^{2}(x)+1;\csc ^{2}(x)=\cot ^{2}(x)+1\]
so
\[2\tan ^{2}(x)+2\cot ^{2}(x)=28;\tan ^{2}(x)+\cot ^{2}(x)=14\]

Answer 5

Also \[\sec ^{2}(x)+\csc ^{2}(x)=16\]

Answer 6

That one goes:
\[\frac {\sin ^{4}(x)+\cos ^{4}(x)}{\sin ^{2}(x)\cos ^{2}(x)}=14\]
The other goes
\[\frac {\sin ^{2}(x)+\cos ^{2}(x)}{\sin ^{2}(x)\cos ^{2}(x)}=16\]

Answer 7

These are all Pythagorean and I am skipping steps, mostly algebraic

Answer 8

yea i got to that step as well

Answer 9

From here you get
\[\sin ^{4}(x)+\cos ^{4}(x)=\frac {7}{8}\]
Again by Pythagorean substitution you get
\[1-2\cos ^{2}(x)+\cos ^{4}(x)=\frac {7}{8}\]
You can use quadratic at this point and you should recognize a double angle formula

Answer 10

\[\cos ^{2}(x)=\frac {2 \pm \sqrt{3}}{4}\]

Answer 11

\[\frac {\sin ^{4}(x)+\cos ^{4}(x)}{\sin ^{2}(x)\cos ^{2}(x)}\div\frac {\sin ^{2}(x)+\cos ^{2}(x)}{\sin ^{2}(x)\cos ^{2}(x)}=7/8\]7/8 should equal to

Answer 12

See the Pythagorean?

Answer 13

oh i get it

Answer 14

thank you

Answer 15

how come sin^4 x is equal to 1 - cos^ x?

Answer 16

\[\sin ^{4}(x)=\sin ^{2}(x)\sin ^{2}(x)=(1-\cos ^{2}(x))(1-\cos ^{2}(x))=1-2\cos ^{2}(x)+\cos ^{4}(x)\]

Answer 17

\[1-2\cos ^{2}(x)+2\cos ^{4}(x)=\frac {7}{8}\]then shouldn't it be

Answer 18

Oops forgot the coefficient of 2 on the fourth power, sorry

Answer 19

thank you so much