Q.Find the principal solution and general solutions of the following:

Part (i), Step 1. Find the reference angle. $\sin\theta=-\dfrac1{\sqrt2}$. Since $\left|-\dfrac1{\sqrt2}\right|=\dfrac1{\sqrt2}=\sin\dfrac\pi4$, the reference angle is $\dfrac\pi4$.

Part (i), Step 2. Fix the quadrant and principal value. Since $\sin\theta<0$, the principal value of sine lies in quadrant IV, i.e. in $\left[-\dfrac\pi2,0\right)$ within the window $\left[-\dfrac\pi2,\dfrac\pi2\right]$. So $\theta=-\dfrac\pi4$, and indeed $\sin\left(-\dfrac\pi4\right)=-\dfrac1{\sqrt2}$. Principal solution: $\theta=-\dfrac\pi4$.

Part (i), Step 3. General solution. With $\alpha=-\dfrac\pi4$, formula (3.13) gives

Part (ii), Step 1. Convert to tangent. $\cot\theta=\sqrt3 \Rightarrow \tan\theta=\dfrac1{\sqrt3}$.

Part (ii), Step 2. Fix the quadrant and principal value. $\dfrac1{\sqrt3}=\tan\dfrac\pi6>0$, so the principal value of tangent lies in quadrant I, within $\left(-\dfrac\pi2,\dfrac\pi2\right)$: $\theta=\dfrac\pi6$. Principal solution: $\theta=\dfrac\pi6$.

Part (ii), Step 3. General solution. With $\alpha=\dfrac\pi6$, the tangent-equation formula gives

Part (iii), Step 1. Find the reference angle. $\tan\theta=-\dfrac1{\sqrt3}$; $\left|-\dfrac1{\sqrt3}\right|=\dfrac1{\sqrt3}=\tan\dfrac\pi6$, so the reference angle is $\dfrac\pi6$.

Part (iii), Step 2. Fix the quadrant and principal value. Since $\tan\theta<0$, the principal value lies in quadrant IV within $\left(-\dfrac\pi2,\dfrac\pi2\right)$: $\theta=-\dfrac\pi6$, and $\tan\left(-\dfrac\pi6\right)=-\dfrac1{\sqrt3}$ ✓. Principal solution: $\theta=-\dfrac\pi6$.

Part (iii), Step 4. General solution. With $\alpha=-\dfrac\pi6$,

Reference angle + quadrant sign to fix the principal value, then the matching general-solution formula

• Reading the reference angle's quadrant from the ORIGINAL ratio (cot, tan) without first converting to the ratio whose window is being used
• Forgetting that a negative sine/tangent principal value sits in quadrant IV (a negative angle), not by adding $\pi$