Question

Please help me do this question​

Answer 1

$x = 90°, 16.3°$

Step-by-step explanation:

We are going to use the identity

$\cos x =\sqrt{1-\sin^2x}$

and substitute this into our expression so we can write

$4\sin x + 3\sqrt{1-\sin^2x} = 4$

$\Rightarrow 3\sqrt{1-\sin^2x} = 4(1 - \sin x)$

Take the square of the equation above to get

$9(1 - \sin^2x) = 16(1 - \sin x)^2$

$\:\:\:\:\:= 16(1 - 2\sin x + \sin^2x)$

Rearranging the terms, we then get

$25\sin^2x - 32\sin x + 7 = 0$

If we let $u = \sin x,$ the above equation becomes

$25u^2 - 32u + 7= 0$

This looks like a quadratic equation whose roots are

$u = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\;\;\;\;= \dfrac{32 \pm \sqrt{32^2 - 4(25)(7)}}{50}$

$\;\;\;\;=\dfrac{32 \pm 18}{50} = 1, 0.28$

We can then write

$\sin x = 1 \:\text{and}\;0.28$

Solving for x, we finally get

$x = 90°\:\text{and}\:16.3°$

Answer 2

Answer:

Step-by-step explanation: