Question

cosθ-11sinθ=6/cosθ
[0°≤θ≤360°]

\(\cos ^{2} θ-11\sin θ\cos θ=6\)
\(1-\sin ^{2}θ-11\sinθ\cosθ=6(\sin ^{2}+\cos ^{2}θ)\)

and then i don't know how to solve.

Answer 1

May be with a change of variable,
\[z=\sin\theta\]
\[1-z^2-11z\sqrt{1-z^2}=6\Rightarrow-11z\sqrt{1-z^2}=5+z^2\]
And solving for z.

Answer 2

my teacher gave me i tip that the second step is need to be continued

Answer 3

we didn't study π with trigonometric

Answer 4

Ok, the trick is completing squares.

Answer 5

\[cosθ-11sinθ=\frac{6}{cos\theta}\]\[cos^2\theta - 11sin\theta \cos\theta = 6\]\[cos^2\theta - 11sin\theta \cos\theta = 6(sin^2\theta+cos^2\theta)\]\[cos^2\theta - 11sin\theta \cos\theta - 6sin^2\theta-6cos^2\theta=0\]\[5cos^2\theta + 11sin\theta \cos\theta+ 6sin^2\theta=0\]Factorize it.
\[(5\cos\theta+6\sin\theta)(cos\theta+\sin\theta)=0\]
I guess you can solve it now

Answer 6

Yes. :)

Answer 7

A very nice hint from your teacher :)

Answer 8

@Callisto thank you, i want to ask is there any program in the calculator fx-50FH of factorizing? i hate factorizing very much ..

Answer 9

You can use the quadratic formula to cheat...
But I usually do it by hand. Practice makes perfect :)

Answer 10

quadratic formula can do this type of question? tests don't have enough time for me to factorize my hand haha