Solve the equation $\sin^2 x + 2 = 3 \cos 2x$ for $0 \leq x < 360$. - Scottish Highers Maths - Question 7 - 2023
Question 7
Solve the equation $\sin^2 x + 2 = 3 \cos 2x$ for $0 \leq x < 360$.
Worked Solution & Example Answer:Solve the equation $\sin^2 x + 2 = 3 \cos 2x$ for $0 \leq x < 360$. - Scottish Highers Maths - Question 7 - 2023
Step 1
Use double angle formula to express equation in terms of $\sin x$
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Answer
We start by using the double angle formula: [ \cos 2x = 1 - 2 \sin^2 x. ] Substituting this into the equation gives us: [ \sin^2 x + 2 = 3(1 - 2 \sin^2 x). ]
Step 2
Arrange in standard quadratic form
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Answer
Rearranging the equation, we get: [ \sin^2 x + 2 = 3 - 6 \sin^2 x, ] which simplifies to [ 7 \sin^2 x - 1 = 0. ]
Step 3
Factorise or use quadratic formula
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Answer
From [ 7 \sin^2 x - 1 = 0, ] we can isolate (\sin^2 x): [ \sin^2 x = \frac{1}{7}. ]
Step 4
Solve for $\sin x$
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Answer
Taking the square root gives: [ \sin x = \pm \frac{1}{\sqrt{7}}. ] This corresponds to: [ \sin x = \frac{1}{\sqrt{7}} \text{ and } \ \sin x = -\frac{1}{\sqrt{7}}. ]
Step 5
Solve for $x$
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Answer
For (\sin x = \frac{1}{\sqrt{7}},) we find the principal values: [ x \approx 19.47^\circ \text{ and } 180 - 19.47 \approx 160.53^\circ. ]
\nFor (\sin x = -\frac{1}{\sqrt{7}},) we find:[ x \approx 360 - 19.47 \approx 340.53^\circ \text{ and } 180 + 19.47 \approx 199.47^\circ. ]
\nThus, the solutions for x are approximately: [ x \approx 19.5^\circ, 160.5^\circ, 199.5^\circ, 340.5^\circ. ]
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