a) Ta có: \(\sin^2\alpha+\cos^2\alpha=1\)
mà \(\sin\alpha=\cos\alpha\)⇒\(2\sin^2\alpha=1\)⇒\(\sin^2\alpha=\frac{1}{2}\)
⇒\(\sin\alpha=\frac{1}{\sqrt{2}}\)⇒ \(\alpha=45\)độ
b) \(2\sin^2\alpha+3\cos^2\alpha=\frac{9}{4}\)
⇔\(2\left(\sin^2\alpha+\cos^2\alpha\right)+\cos^2\alpha=\frac{9}{4}\)⇒\(\cos^2\alpha=\frac{1}{4}\)
⇔\(\cos\alpha=\frac{1}{2}\)⇒\(\alpha=30\) dộ