Put \(c=\cos(x/2)\), where \(0\leq x/2\leq\pi\). Using \(\cos x=2c^2-1\), \(2(2c^2-1)-c=1\), so \(4c^2-c-3=0=(c-1)(4c+3)\). Hence \(c=1\) or \(-3/4\). Therefore \(\boxed{x=0,\ 2\cos^{-1}(-3/4)=4.84\text{ radians}}\) (3 s.f.).

(a) Use \(\sin2\theta=2\sin\theta\cos\theta\) and \(\cos2\theta=\cos^2\theta-\sin^2\theta\), then move the right-hand term to the left, obtaining the stated expression.

(b) Factor: \((\cos\theta-\sin\theta)(\cos\theta+3\sin\theta)=0\). Thus \(\tan\theta=1\) or \(-1/3\). In the required interval, \(\boxed{\theta=45^\circ,\ 161.6^\circ}\) (1 d.p.). The excluded case \(\cos\theta=0\) does not solve the original equation.

(a) Expansion gives \(4\cos x+\sqrt3\sin x\). Comparing with \(R\cos\alpha\cos x+R\sin\alpha\sin x\), \(R\cos\alpha=4,\ R\sin\alpha=\sqrt3\). Thus \(\boxed{R=\sqrt{19},\ \alpha=\tan^{-1}(\sqrt3/4)=23.41^\circ}\).

(b) \(\cos(2\theta-\alpha)=2.5/\sqrt{19}\). Let \(\beta=\cos^{-1}(2.5/\sqrt{19})\). In \(-\alpha<2\theta-\alpha<360^\circ-\alpha\), the solutions are \(\beta,360^\circ-\beta\). Therefore \(\theta=(\alpha+\beta)/2\) or \((\alpha+360^\circ-\beta)/2\), giving \(\boxed{\theta=39.2^\circ,\ 164.2^\circ}\) (1 d.p.).

(a) Comparing coefficients gives \(R\cos\alpha=12\), \(R\sin\alpha=5\). Hence \(\boxed{R=13,\ \alpha=\tan^{-1}(5/12)=0.395\text{ radians}}\) (3 s.f.).

(b) \(\cos(2x-\alpha)=6/13\). Put \(\beta=\cos^{-1}(6/13)\). The solutions in the required range are \(2x-\alpha=\beta\) and \(2\pi-\beta\). Thus \(\boxed{x=(\alpha+\beta)/2=0.743,\ (\alpha+2\pi-\beta)/2=2.79}\) radians (3 s.f.).

By the product rule, \[\frac{dy}{dx}=\cos x\sin2x+2\sin x\cos2x.\] Applying double-angle identities, \[\frac{dy}{dx}=2\sin x\cos^2x+2\sin x(\cos^2x-\sin^2x)=2\sin x(3\cos^2x-1).\] In the given open interval, \(\sin x>0\) and \(\cos x>0\). Thus a stationary point satisfies \[\cos x=1/\sqrt3,\qquad x=\cos^{-1}(1/\sqrt3)\approx\boxed{0.955}\text{ radians}.\]

(a) By the product and chain rules, \[\frac{dy}{dx}=e^{-4x}\sec^2x-4e^{-4x}\tan x.\] Factor out \(e^{-4x}\sec^2x\) and use \(\tan x/\sec^2x=\sin x\cos x=\tfrac12\sin2x\): \[\boxed{\frac{dy}{dx}=\sec^2x(1-2\sin2x)e^{-4x}}.\] Thus \(a=1\), \(b=-2\).

(b) The exponential and squared secant are positive in the interval, so \[1-2\sin2x=0,\qquad\sin2x=1/2.\] In \(0\le2x<\pi\), the solutions are \(2x=\pi/6,5\pi/6\). Hence \[\boxed{x=\frac\pi{12},\quad\frac{5\pi}{12}}.\]

Apply the product and chain rules: \[\frac{dy}{dx}=-3\cos^2x\sin x\sqrt{\sin x}+\frac{\cos^4x}{2\sqrt{\sin x}}.\] In the stated interval the sine and cosine are positive. Multiplying the zero-derivative equation by \(2\sqrt{\sin x}\) gives \[-6\cos^2x\sin^2x+\cos^4x=0.\] Divide by \(\cos^2x\): \[\cos^2x=6\sin^2x,\qquad\tan^2x=1/6.\] Therefore \[\boxed{x=\tan^{-1}(1/\sqrt6)\approx0.388}\quad\text{(radians)}.\]

(a) Quotient differentiation gives \[\frac{dy}{dx}=\frac{3\cos3x(2-\cos3x)-3\sin3x(3+\sin3x)}{(2-\cos3x)^2}=\frac{3(2\cos3x-1-3\sin3x)}{(2-\cos3x)^2}.\] Setting this equal to \(-3/2\) at \(x=a\), and writing \(c=\cos3a,s=\sin3a\), gives \[2(2c-1-3s)=-(2-c)^2\iff c^2-6s+2=0.\] Since \(c^2=1-s^2\), \(\boxed{s^2+6s-3=0}\), as required.

(b) \(s=-3\pm2\sqrt3\). Only \(s=2\sqrt3-3\) lies in \([-1,1]\). Let \(u=\arcsin(2\sqrt3-3)\). Since \(0\le3a\le3\pi\), \[3a=u,\ \pi-u,\ 2\pi+u,\ 3\pi-u.\] Thus \(\boxed{a=0.161,\ 0.886,\ 2.26,\ 2.98}\) radians (3 significant figures).

(a) By the quotient rule, \[y\prime=\frac{(5+2\sin x)\sec^2x-2\cos x\tan x}{(5+2\sin x)^2}.\] Multiply the numerator by \(\cos^2x\): \[5+2\sin x-2\sin x\cos^2x=5+2\sin^3x.\] Therefore \[y\prime=\frac{5+2\sin^3x}{(1-\sin^2x)(5+2\sin x)^2},\quad a=5,\ b=2.\]

(b) Since \(-1\le\sin x\le1\), the numerator is at least 3, so cannot vanish. The denominator is positive wherever the curve is defined; points with \(\cos x=0\) are excluded. Hence there are no stationary points.

(a) Differentiate in the interior: \[y\prime=-\sin x\sqrt{\sin2x}+\frac{\cos x\cos2x}{\sqrt{\sin2x}}.\] At a stationary point, \[-\sin a\sin2a+\cos a\cos2a=0\Rightarrow\cos3a=0.\] The sole interior solution is \(a=\pi/6\); \(a=\pi/2\) is an endpoint outside this derivative calculation. The endpoints have \(y=0\) and the curve is positive inside, so \[a=\frac\pi6.\]

(b) \[V=\pi\int_0^{\pi/2}y^2\,dx=\pi\int_0^{\pi/2}\cos^2x\sin2x\,dx=2\pi\int_0^{\pi/2}\cos^3x\sin x\,dx.\] Thus \[V=\left[-\frac\pi2\cos^4x\right]_0^{\pi/2}=\frac\pi2.\]

(a) Rewrite \(y=10\sin x\cos^3x\). Differentiate: \[y\prime=10\cos^4x-30\sin^2x\cos^2x=10\cos^2x(\cos^2x-3\sin^2x).\] In the interior, stationarity requires \[1-4\sin^2x=0\Rightarrow\sin x=\frac12\Rightarrow x=\frac\pi6.\] The derivative changes from positive to negative here and both endpoints have zero height, so this is the maximum.

(b) With \(u=\cos x\), \(du=-\sin x\,dx\); the limits are 1 and \(1/\sqrt2\). Thus \[\text{area}=-10\int_1^{1/\sqrt2}u^3\,du=\left[-\frac52u^4\right]_1^{1/\sqrt2}=\frac52\left(1-\frac14\right)=\frac{15}8.\]

(a) Expand both sides: \(2\cos x\sin(\pi/6)=-2\sin x\sin(\pi/3)\), so \(\cos x=-\sqrt3\sin x\). Since \(\cos x=0\) would not satisfy the equation, division is valid and \(\boxed{\tan x=-1/\sqrt3}\).

(b) Tangent is negative in quadrants II and IV, with reference angle \(\pi/6\). Thus \(\boxed{x=5\pi/6,\ 11\pi/6}\).

(a) \(\cos(2\theta+\theta)=\cos2\theta\cos\theta-\sin2\theta\sin\theta=(2\cos^2\theta-1)\cos\theta-2\sin^2\theta\cos\theta=4\cos^3\theta-3\cos\theta\).

(b) Let \(c=\cos\theta\). Using the identity and \(\cos2\theta=2c^2-1\) gives \(4c^3-3c+c(2c^2-1)=c^2\), so \(c(6c^2-c-4)=0\). Hence \(c=0\) or \(c=(1\pm\sqrt{97})/12\). The three solutions are \(\boxed{\theta=25.3^\circ,\ 90^\circ,\ 137.5^\circ}\) (1 d.p.).

(a) Put \(s=\sin^2\theta\). Since the original cotangent requires \(s\ne0\), \((1-s)/s+2(1-2s)=4\). Multiplying by \(s\) and rearranging gives \(4s^2+3s-1=0\), as required.

(b) \((4s-1)(s+1)=0\). Reject \(s=-1\), leaving \(\sin^2\theta=1/4\). Thus \(\boxed{\theta=30^\circ,\ 150^\circ,\ 210^\circ,\ 330^\circ}\).