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Please help me do this question​

Please help me do this question​-example-1
User Nkron
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2 Answers

14 votes
14 votes


x = 90°, 16.3°

Explanation:

We are going to use the identity


\cos x =√(1-\sin^2x)

and substitute this into our expression so we can write


4\sin x + 3√(1-\sin^2x) = 4


\Rightarrow 3√(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 = (-b \pm √(b^2 - 4ac))/(2a)


\;\;\;\;= (32 \pm √(32^2 - 4(25)(7)))/(50)


\;\;\;\;=(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°

User Midhun Darvin
by
2.5k points
15 votes
15 votes

Answer:

Explanation:

User Todd Freed
by
3.3k points
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