Moderate -0.3 This is a standard trigonometric equation requiring the identity cos²x = 1 - sin²x to convert to a quadratic in sin x, then solving the quadratic and finding angles in a given range. It's slightly easier than average as it follows a well-practiced procedure with no conceptual surprises, though it requires careful execution across multiple steps.
4 In this question you must show detailed reasoning.
Solve the equation \(6 \cos ^ { 2 } x + \sin x = 5\), giving all the roots in the interval \(- 180 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\).
Collects terms and attempts to solve their quadratic
\(\sin x = \frac{1}{2}, -\frac{1}{3}\)
A1
soi. May be BC. FT their quadratic
When \(\sin x = \frac{1}{2}\): \(x = 30°, 150°\)
A1
At least one correct root in the interval for either value of \(\sin x\). FT their valid value for \(\sin x\)
When \(\sin x = -\frac{1}{3}\): \(x = -19.5°, -160.5°\)
A1
All roots seen from complete working – no extras in the range. FT their other valid value for \(\sin x\). Ignore additional answers outside the range.
## Question 4:
$6(1 - \sin^2 x) + \sin x = 5$ | M1 | Uses the identity $\sin^2 x + \cos^2 x = 1$
$6\sin^2 x - \sin x - 1 = 0$ | M1 | Collects terms and attempts to solve their quadratic
$\sin x = \frac{1}{2}, -\frac{1}{3}$ | A1 | soi. May be BC. FT their quadratic
When $\sin x = \frac{1}{2}$: $x = 30°, 150°$ | A1 | At least one correct root in the interval for either value of $\sin x$. FT their valid value for $\sin x$
When $\sin x = -\frac{1}{3}$: $x = -19.5°, -160.5°$ | A1 | All roots seen from complete working – no extras in the range. FT their other valid value for $\sin x$. Ignore additional answers outside the range.
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4 In this question you must show detailed reasoning.\\
Solve the equation $6 \cos ^ { 2 } x + \sin x = 5$, giving all the roots in the interval $- 180 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.
\hfill \mbox{\textit{OCR MEI AS Paper 1 2023 Q4 [5]}}