Show that the equation
3sin2θ−2cos2θ=1 3 \sin^2 \theta - 2 \cos^2 \theta = 1 3sin2θ−2cos2θ=1can be written as
5sin2θ=3. 5 \sin^2 \theta = 3. 5sin2θ=3.Hence solve, for 0∘≤θ<360∘0^\circ \leq \theta < 360^\circ0∘≤θ<360∘, the equation
3sin2θ−2cos2θ=1, 3 \sin^2 \theta - 2 \cos^2 \theta = 1, 3sin2θ−2cos2θ=1,giving your answer to 1 decimal place.
10 exam-style questions on Edexcel A Level Old Maths Trigonometry. Each one has a worked solution and a mark scheme showing where the marks go.