Moderate -0.3 This is a standard trigonometric equation requiring substitution of sin²θ = 1 - cos²θ to form a quadratic in cos θ, then solving. It's slightly easier than average because the method is routine and well-practiced, though students must remember the Pythagorean identity and handle two solutions in the given range.
substitution of \(\sin^2\theta = 1 - \cos^2\theta\)
M1
soi
\(-5\cos^2\theta = \cos\theta\)
A1
or better
\(\theta = 90\) and \(270\)
A1
\(102\)
A1
accept \(101.5(\ldots)\) and \(258.(46\ldots)\) rounded to 3 or more sf; if M0, allow B1 for both of 90 and 270
\(258\)
A1
\(101\) *and* \(259\)
SC1
and B1 for 102 and B1 for 258 (to 3 or more sf)
**Question 8:**
| Answer | Mark | Guidance |
|--------|------|----------|
| substitution of $\sin^2\theta = 1 - \cos^2\theta$ | **M1** | soi | |
| $-5\cos^2\theta = \cos\theta$ | **A1** | or better | |
| $\theta = 90$ and $270$ | **A1** | | if the 4 correct values are presented, ignore any extra values which are outside the required range, but apply a penalty of minus 1 for extra values in the range |
| $102$ | **A1** | accept $101.5(\ldots)$ and $258.(46\ldots)$ rounded to 3 or more sf; if **M0**, allow **B1** for both of 90 and 270 | |
| $258$ | **A1** | | |
| $101$ *and* $259$ | **SC1** | and **B1** for 102 and **B1** for 258 (to 3 or more sf) | if given in radians deduct 1 mark from total awarded $(1.57, 1.77, 4.51, 4.71)$ |
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