Question

4(sin x)^2 + 2 (cos x)^2 = 3. How do I solve the equation in the interval [0, 2pi) ?

Answer 1

you can replace
\[\cos^2(x)\] by \[1-\sin^2(x)\] to get an equation only in sine

Answer 2

re-write as:
\[2\sin ^{2}x +2\cos ^{2}x + 2\sin ^{2}x=3\]
\[2(\sin ^{2}x+\cos ^{2}x)+2\sin ^{2}x=3\]
can you pick up from that?

Answer 3

\[2+2\sin ^{2}x=3\]
\[2\sin ^{2}x = 1\]
finish it :)

Answer 4

\[sinx=+/- 1/\sqrt{2}\]
so, x=+/- pi/4

Answer 5

how did you get rid of the cosine?

Answer 6

sin^2 + cos^2 = 1

Answer 7

ohh okay i see now

Answer 8

thank youuuu :)

Answer 9

welcome!

Answer 10

Please click: 'Good Answer' button on the right... it will remove this problem from unanswered :)